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Electrostatics - Practice Questions (39)

Question 1: 1. As shown in the figure, the electric field strength of $E$ in a uniform electric field, fixed a c...

1. As shown in the figure, the electric field strength of $E$ in a uniform electric field, fixed a charged metal ball P, at this time $M$ point electric field strength is $0 , ~ P , M$ line is parallel to the electric field line. A metal ball Q, which is identical to ball P and is not charged, is moved to infinity after contact with ball P. The electric field at $M$ is ( ) ![](/images/questions/phys-electrostatics/image-001.jpg)

  • A. A. $E$
  • B. B. $\frac { E } { 2 }$
  • C. C. $\frac { E } { 3 }$
  • D. D. $\frac { 2 E } { 3 }$

Answer: B

Solution: According to the meaning of the question can be seen, the ball P in $M$ point produced by the electric field strength size of $E$, the direction of the electric field strength and the direction of the opposite direction of the electric field strength of the uniform electric field, the metal ball Q and the metal ball P contact, the charge of the ball P become half of the original, then the ball P in $M$ point produces the electric field strength direction remains unchanged, the size also becomes half of the original, that is, $\frac { E } { 2 }$. FORMULA_3]] point produced by the electric field strength direction is unchanged, the size of the original half, that is, $\frac { E } { 2 }$, then $M$ point of the combined field size is $$ E ^ { \prime } = E - \frac { E } { 2 } = \frac { E } { 2 } $$

Question 2: 2. The interaction force between two point charges in a vacuum is $F$, if the distance between the c...

2. The interaction force between two point charges in a vacuum is $F$, if the distance between the charges remains unchanged, the charge of one of the charges is changed to 3 times the original, then the interaction force at this time becomes ( )

  • A. A. $4 F$
  • B. B. $3 F$
  • C. C. $\frac { F } { 3 }$
  • D. D. $9 F$

Answer: B

Solution: The Coulomb force before the change in charge is $$ F = k \frac { q _ { 1 } q _ { 2 } } { r ^ { 2 } } $$ The Coulomb force after the change is $$ F ^ { \prime } = k \frac { 3 q _ { 1 } q _ { 2 } } { r ^ { 2 } } = 3 F $$

Question 3: 3. Regarding the potential in an electrostatic field, the following statements are correct ( )

3. Regarding the potential in an electrostatic field, the following statements are correct ( )

  • A. A. The potential of an uncharged conductor must be equal to zero.
  • B. B. The potential at a point where the electric field strength is zero must be zero
  • C. C. The potential must decrease along the electric field lines.
  • D. D. The potentials of points on the same electric field line must be equal.

Answer: C

Solution: AB. In the electrostatic field, the potential is relative, and zero potential reference surface selection, so AB error; CD. Along the direction of the electric field lines of potential decrease, the same electric field lines on the points, the potential must not be equal, so C is correct, D is wrong.

Question 4: 4. The following statements about electric field strength and potential are correct ( ).

4. The following statements about electric field strength and potential are correct ( ).

  • A. A. Where the potential is high, the electric field is strong.
  • B. B. The electric field strength must be zero where the potential is zero.
  • C. C. All points in a uniform electric field have equal potentials
  • D. D. The potential always decreases along the direction of the electric field lines

Answer: D

Solution: A. The field strength is related to the sparseness of the electric field lines, and the electric field strength may be smaller in the place of high potential, A error; B. The field strength is not necessarily 0 where the potential can be 0, B is wrong; C. uniform electric field, the field strength is equal everywhere, along the direction of the electric field lines, the potential decreases, the potential are not equal, so C error; D. Along the direction of the electric field lines, the potential decreases, D is correct.

Question 5: 5. The force between two electrically charged metal spheres $A$, B ( )

5. The force between two electrically charged metal spheres $A$, B ( )

  • A. A. It must be gravity.
  • B. B. It must be repulsion.
  • C. C. If they have the same charge, they must be gravitational.
  • D. D. indeterminate

Answer: D

Solution: If the two metal balls with the same kind of charge, between the repulsive force, if the two metal balls with different kinds of charge, between the gravitational force, the question is not enough conditions, so can not be determined, D is correct.

Question 6: 6. As shown in the figure, there are four points on the circle with radius $r$ exactly forming a squ...

6. As shown in the figure, there are four points on the circle with radius $r$ exactly forming a square, and there are point charges fixed on the corresponding points. The charge of the point charge has been marked in the figure, then the center of the circle $O$ at the size of the electric field strength is (electrostatic force constant for $k$) ( ![](/images/questions/phys-electrostatics/image-002.jpg)

  • A. A. $\frac { 5 k Q } { r ^ { 2 } }$
  • B. B. $\frac { 4 k Q } { r ^ { 2 } }$
  • C. C. $\frac { 3 k Q } { r ^ { 2 } }$
  • D. D. $\frac { k Q } { r ^ { 2 } }$

Answer: D

Solution: ground $$ E = \frac { k Q } { r ^ { 2 } } $$ It can be seen that the field strengths generated by $- Q$ and -$Q$ at point $O$ are equal in magnitude, opposite in direction, and cancel each other, and the field strength generated by $2 Q$ at point and $Q$ at $O$ are in the opposite direction, and the resulting field strength is $$ E = \frac { 2 k Q } { r ^ { 2 } } - \frac { k Q } { r ^ { 2 } } = \frac { k Q } { r ^ { 2 } } $$ D is correct.

Question 7: 7. The following statements are correct ( )

7. The following statements are correct ( )

  • A. A. Newton discovered the law of gravity and calculated the gravitational constant through experimental measurements $G$
  • B. B. Cavendish used a Coulomb torsion scale to cleverly realize his study of the law of the force of interaction between charges
  • C. C. German astronomer Kepler discovered the law of gravity and formulated Newton's three laws
  • D. D. The French physicist Coulomb discovered the law of interaction between charges, Coulomb's law, using a torsion scale experiment, and measured the value of the electrostatic force constant K.

Answer: D

Solution: Newton discovered the law of universal gravitation, Cavendish measured the gravitational constant $G$ by means of a torsion scale experiment, so $A$ is wrong; the law of universal gravitation and Newton's three laws are summarized in Newton's discovery, so C is wrong; Coulomb discovered the law of the interaction of electric charges, Coulomb's law, and measured the electrostatic force constant, so B is wrong, D is correct; Coulomb discovered the law of the interaction of electric charges, Coulomb's law, and measured the electrostatic force constant, so B is wrong, D is correct. Coulomb discovered the law of interaction between electric charges, Coulomb's law, and measured the electrostatic force constant, so B is wrong and D is correct.

Question 8: 8. A point charge $O$ in a vacuum has two points $A , B$ at equal distances to the left and right, t...

8. A point charge $O$ in a vacuum has two points $A , B$ at equal distances to the left and right, then ( ) ![](/images/questions/phys-electrostatics/image-003.jpg)

  • A. A. The electric field strength at point A is the same as that at point $B$.
  • B. B. The potential at point A is the same as the potential at point B.
  • C. C. Test charge-$q$ has a greater potential energy at point $A$ than at point $B$
  • D. D. Negative work done by the electric field force in moving the test charge $- q$ from the point $A$ to the point $B$

Answer: B

Solution: A. According to the symmetry of the electric field of the point charge, it can be seen that the electric field strength of the $A$ point and the electric field strength of the $B$ point are of the same magnitude, and the electric field strength of the $B$ point is the same. The direction is opposite. Therefore, A is wrong; B. According to the characteristics of the equipotential surface of the electric field of a point charge, it can be seen that the electric potential at point $A$ is the same as that at point $B$. Therefore, $B$ is correct; C. According to the defining equation of potential energy, there are $$ E _ { P } = q \varphi , \varphi _ { \mathrm { A } } = \varphi _ { \mathrm { B } } $$ The potential energy of the test charge $- q$ at $A$ is equal to the potential energy at $B$. Therefore, C is wrong; D. According to the formula of work done by the electric field force, there are $$ W = q U , U _ { \mathrm { AB } } = \varphi _ { \mathrm { A } } - \varphi _ { \mathrm { B } } = 0 $$ The electric field force does no work throughout the process of moving the test charge-$q$ from point $A$ to point $B$. Therefore, D is wrong. The force of the electric field does not do any work during the whole process.

Question 9: 9. It is known that the electric field strength at a point in the electric field is $1 \times 10 ^ {...

9. It is known that the electric field strength at a point in the electric field is $1 \times 10 ^ { 5 } \mathrm {~N} / \mathrm { C }$, then the electric field force on a point charge with a charge of $1.6 \times 10 ^ { - 19 } \mathrm { C }$ at the point is the magnitude of the electric field force ( )

  • A. A. $1.6 \times 10 ^ { - 14 } \mathrm {~N}$
  • B. B. $1.6 \times 10 ^ { - 15 } \mathrm {~N}$
  • C. C. $1.6 \times 10 ^ { - 13 } \mathrm {~N}$
  • D. D. $1.6 \times 10 ^ { - 16 } \mathrm {~N}$

Answer: A

Solution: ground $$ F = E q $$ It is known that the magnitude of the electric field force on a point charge at that point is $$ F = 1 \times 10 ^ { 5 } \times 1.6 \times 10 ^ { - 19 } \mathrm {~N} = 1.6 \times 10 ^ { - 14 } \mathrm {~N} $$

Question 10: 11. The following narratives are correct

11. The following narratives are correct

  • A. A. Charge cannot be created, but it can be destroyed
  • B. B. An object has a negative charge of $1.6 \times 10 ^ { - 9 } \mathrm { C }$, which is due to the fact that it has gained $1.0 \times 10 ^ { 10 }$ electrons
  • C. C. Only spherical charged bodies of small size are called point charges
  • D. D. A meta-charge is a unit charge

Answer: B

Solution: Charge will not be created out of thin air or disappear out of thin air, i.e., charge cannot be created or destroyed, so A is wrong; an object is negatively charged with $1.6 \times 10 ^ { - 9 } \mathrm { C }$, which is due to the fact that it has gained $1.0 \times 10 ^ { 10 }$ electrons, so B is correct; when the shape of a charged body has negligible influence on the interaction forces between them, the This charged body can be regarded as a point charge, with the charge The size of the charge has nothing to do with, so C is wrong; meta-charge refers to the charged amount of protons or electrons, not protons, electrons, not the unit charge, so D is wrong. So B is correct, ACD is wrong.

Question 11: 12. There are two stationary point charges in a vacuum, the force between them is $F$, if their char...

12. There are two stationary point charges in a vacuum, the force between them is $F$, if their charge is increased to 3 times the original, the distance is reduced to the original $\frac { 1 } { 3 }$, the interaction force between them becomes

  • A. A. $\frac { F } { 9 }$
  • B. B. $F$
  • C. C. $9 F$
  • D. D. $81 F$

Answer: D

Solution: The force between two stationary point charges in a vacuum satisfies the conditions of Coulomb's law, so according to Coulomb's law, the Coulomb force between them before the change in charge and distance is $$ F = k \frac { Q _ { 1 } Q _ { 2 } } { r ^ { 2 } } $$ After their charges are both increased to 3 times their original amount and the distance is reduced to $\frac { 1 } { 3 }$ $$ F ^ { \prime } = k \frac { 3 Q _ { 1 } \cdot 3 Q _ { 2 } } { \left( \frac { r } { 3 } \right) ^ { 2 } } = 81 k \frac { Q _ { 1 } Q _ { 2 } } { r ^ { 2 } } = 81 F $$

Question 12: 13. A test charge is placed at point $c$ near a point charge $Q$, and test charges of different char...

13. A test charge is placed at point $c$ near a point charge $Q$, and test charges of different charges are placed successively at point $c$, and the electric field force on the test charge is different. The following images describe the relationship between the magnitude $F$ of the force on the test charge and its charge $q$, and each of $A , B$ represents a set of $F , q$ data.

  • A. A. ![](/images/questions/phys-electrostatics/image-001.jpg)
  • B. B. ![](/images/questions/phys-electrostatics/image-002.jpg)
  • C. C. ![](/images/questions/phys-electrostatics/image-003.jpg)
  • D. D. ![](/images/questions/phys-electrostatics/image-004.jpg)

Answer: C

Solution: The magnitude of the electric field force on the test charge is $$ F = \frac { k Q } { r ^ { 2 } } q $$ The position of the test charge has been placed at the point $c$ according to the $$ E = \frac { k Q } { r ^ { 2 } } $$ So the electric field strength of the test charge is of the same magnitude, and the slope of the image indicates the magnitude of the field strength, so the slope of the line connecting the point $A$ to the origin is the same as that of the line connecting the point $B$ to the origin and is the image past the origin.

Question 13: 14. As shown in the figure shows the distribution of electric field lines formed by point charge $a ...

14. As shown in the figure shows the distribution of electric field lines formed by point charge $a , b$, the following statement is correct () ![](/images/questions/phys-electrostatics/image-004.jpg)

  • A. A. $A$ point potential equals $B$ point potential
  • B. B. $a , b$ is a homogeneous charge
  • C. C. $A$ point field is stronger than $B$ point field.
  • D. D. The $A$ point potential is higher than the $B$ point potential.

Answer: D

Solution: ABD. The electric field lines start from positive charge or infinity and terminate at infinity or negative charge, so $a$ should be positive charge, $b$ is negative charge, and the potential decreases along the direction of the electric field lines, so that the potential at point $A$ is higher than that at point $B$. FORMULA_3]] is a negative charge. D is correct; C. The denser the electric field lines, the stronger the field, and the field at $A$ is smaller than the field at $B$.C is incorrect.

Question 14: 15. The following statements are correct ()

15. The following statements are correct ()

  • A. A. If the change in momentum of an object is greater, then the impulse of the combined external force on it must be greater
  • B. B. The impulse of a larger force must be greater than the impulse of a smaller force
  • C. C. The rate of an object with more momentum must be greater than the rate of an object with less momentum
  • D. D. A small enough charge is a point charge

Answer: A

Solution: A. According to the Momentum Theorem $I _ { \text {合 } } = \Delta p$, the impulse of the force on an object is equal to the change in momentum, so the greater the impulse of the force on an object, the greater the change in momentum, so A is correct; B. According to the definition of impulse $I = F t$, it can be seen that the impulse of a large force is not necessarily larger than that of a small force, and it is also related to the time of action of the force, so help B is wrong; C. According to the definition of momentum $p = m v$, it can be known that the rate of an object with large momentum is not necessarily larger than the rate of an object with small momentum, but also related to the mass of the object, so C is wrong; D. can be seen as a point charge, depends on the shape of the charge, the volume of the problem analyzed whether the effect of negligible, rather than small enough charge is a point charge, so D error;

Question 15: 16. The following statements are correct ( ) is directly proportional and inversely proportional...

16. The following statements are correct ( ) is directly proportional and inversely proportional to the charge $q$ of the test charge

  • A. A. Coulomb's study of the force between charges in relation to distance and charge applies the controlled variable method
  • B. B. The $d$ in equation $E = \frac { U } { d }$ is the distance between two points in a uniform electric field with a potential difference of $U$.
  • C. C. From the formula $E = \frac { F } { q }$, the field strength at a point in an electric field is related to the magnitude of the electric field force $F$ on a test charge placed at that point
  • D. D. According to $B = \frac { F } { I L }$ it is known that a small piece of energized wire is not subject to magnetic force at a certain place, then the magnetic induction must be zero at that place

Answer: A

Solution: A. Coulomb's study of the relationship between the force between charges and the distance, the amount of charge when the first control charge charge unchanged to study the relationship between the force and the distance, and then control the distance unchanged to study the relationship between the force and the amount of charge, this method is the control variable method, A correct; B. formula $E = \frac { U } { d }$ in the $d$ for the uniform electric field potential difference for $U$ of the two points along the direction of the electric field lines on the distance, B is wrong; C. the formula $E = \frac { F } { q }$ is the definition of the strength of the electric field, [[]] is the definition of the electric field strength, [[]] is the definition of the electric field strength. INLINE_FORMULA_4]] depends on the electric field itself, and has nothing to do with the magnitude of the electric field force $F$ on the test charge placed at the point, or the amount of electricity $q$ on the test charge; D. When a small piece of energized wire is placed parallel to the direction of the magnetic field, it is not subject to the magnetic field force, but the magnetic induction is not necessarily zero at that location. d. wrong.

Question 16: 17. With respect to the understanding of Coulomb's law, the following statement is correct

17. With respect to the understanding of Coulomb's law, the following statement is correct

  • A. A. For the calculation of the electrostatic force between any charged bodies, Coulomb's law formula can be used
  • B. B. The electrostatic force between two point charges, whether in a vacuum or in a medium, must be equal in magnitude and opposite in direction
  • C. C. As long as the electrostatic force between point charges is calculated, the Coulomb's law formula can be used
  • D. D. A rubbed rubber mallet attracts confetti, which means that the confetti must be positively charged.

Answer: B

Solution: AC. Coulomb's law applies to the calculation of electrostatic force between stationary point charges in vacuum, so AC is wrong; B. The electrostatic force between two point charges is the relationship between the action force and reaction force, so whether in a vacuum or in the medium, must be equal in size and opposite direction, so B is correct; D. friction over the rubber rod to attract confetti, confetti positively charged or uncharged can be, so D error.

Question 17: 18. Which of the following statements about point charges is true?

18. Which of the following statements about point charges is true?

  • A. A. Point charges are real.
  • B. B. Larger charged bodies that cannot be seen as point charges
  • C. C. Point charge is not an ideal model
  • D. D. Whether a charged body can be seen as a point charge is not a function of the absolute value of its size, but of whether its shape and size have a negligible effect on the interaction force

Answer: D

Solution: AC. Point charge is not real, it is an idealized model, so AC is wrong; BD. Whether it can be regarded as a point charge depends on the shape and size of the charged body and the charge distribution can be ignored in the problem under study, and it has nothing to do with the size of the charged body, so B is wrong and D is correct.

Question 18: 19. There are two differently charged identical metal spheres A and B (which can be regarded as poin...

19. There are two differently charged identical metal spheres A and B (which can be regarded as point charges) in a vacuum, with charge sizes $q$ and $3 q$ respectively, and the Coulomb force between them is $F$ when the distance between them is $F$, and now the two spheres A and B touch and then separate and return to their original positions, then the Coulomb force between them is $r$. FORMULA_3]], now the two balls A and B are separated after contact and put back to the original position, then the magnitude of the Coulomb force between them is

  • A. A. $\frac { F } { 3 }$
  • B. B. $\frac { 2 F } { 3 }$
  • C. C. $F$
  • D. D. $\frac { 4 F } { 3 }$

Answer: A

Solution: The magnitude of the Coulomb force between spheres A and B, when separated by $r$, is $$ F = k \frac { q \cdot 3 q } { r ^ { 2 } } = \frac { 3 k q ^ { 2 } } { r ^ { 2 } } $$ When the $A , B$ spheres are brought into contact and then separated, the charged quantity of the spheres will be $$ q ^ { \prime } = \frac { 3 q - q } { 2 } = q $$ The Coulomb force between A and B, when the two balls are put in contact and then separated and put back to their original positions, is $$ F ^ { \prime } = k \frac { q ^ { \prime 2 } } { r ^ { 2 } } = \frac { k q ^ { 2 } } { r ^ { 2 } } = \frac { F } { 3 } $$

Question 19: 20. As shown in the figure, $A , B$ is an electric field line on the two points, if the release of a...

20. As shown in the figure, $A , B$ is an electric field line on the two points, if the release of an initial speed of zero at a certain point of the electron, the electron is only by the action of the electric field force, and along the electric field line from the $A$ movement to the $B$, and its speed with the time change of the law shown in the figure. Then ![](/images/questions/phys-electrostatics/image-005.jpg) ![](/images/questions/phys-electrostatics/image-006.jpg)

  • A. A. Electric field force $\mathrm { F } _ { \mathrm { A } } < \mathrm { F } _ { \mathrm { B } }$
  • B. B. Electric field strength $\mathrm { E } _ { \mathrm { A } } = \mathrm { E } _ { \mathrm { B } }$
  • C. C. Potential $U _ { A } > U _ { B }$
  • D. D. Potential energy $E _ { p } A < E _ { p } B$

Answer: A

Solution: C. The electron accelerates, and the force is to the right, so the electric field strength direction is from $B$ to $A , A$ and the potential is lower than that at $B$, so C is wrong; AB. The slope of the velocity-time image indicates the acceleration, from the figure, from $A$ to $B$ process, the acceleration becomes larger, then the electric field force becomes larger, the field strength becomes larger, so A is correct, B is wrong; D. The electrons have more potential energy at the lower potential, so the potential energy of the $A$ point is larger, so $D$ is wrong.

Question 20: 21. The following description of the strength of the electric field is correct ( )

21. The following description of the strength of the electric field is correct ( )

  • A. A. The field strength at a point in an electric field is numerically equal to the electric field force on a unit charge
  • B. B. The field strength at a point in an electric field is proportional to the electric field force on the test charge at that point
  • C. C. The direction of the field strength at a point in an electric field is the direction of the electric field force on the test charge at that point
  • D. D. The field strength at a point in an electric field is related to the presence or absence of a test charge at that point

Answer: A

Solution: A. From the defining equation of electric field strength $$ E = \frac { F } { q } $$ The field strength at a point in the electric field is numerically equal to the electric field force on the unit charge, so A is correct; B. The size of the field strength at a point in the electric field is determined by the electric field itself, unrelated to the test charge, so B is wrong; C. Physics states that the direction of the field strength at a point in the electric field is the same as the direction of the electrostatic force on the positive charge at that point, so C is wrong; D. The field strength at a point in the electric field indicates the strength and direction of the electric field itself, with or without the test charge, so D error.

Question 21: 22. The correct statement about electric potential and potential difference is

22. The correct statement about electric potential and potential difference is

  • A. A. Potential and potential difference are both positive and negative, they are vectors
  • B. B. Potential is vector, potential difference is scalar
  • C. C. The units of potential and potential difference are both V
  • D. D. Potential and potential difference are relative and related to the position of zero potential

Answer: C

Solution: A B. Potential and potential difference are both positive and negative, and are scalar quantities, so AB is wrong; C. The unit of potential and potential difference is $V$, so $C$ is correct; D. The potential is relative to the zero potential position, and the potential difference is absolute independent of the zero potential position, so D is wrong. The potential difference is absolute and independent of the zero potential position, so D is wrong.

Question 22: 23. The figure shows the distribution of electric field lines for a point charge Q. $\mathrm { A } ,...

23. The figure shows the distribution of electric field lines for a point charge Q. $\mathrm { A } , \mathrm {~B}$ is two points on a particular line, then () ![](/images/questions/phys-electrostatics/image-007.jpg)

  • A. A. Q Negatively charged
  • B. B. Point A has a higher potential than point $B$.
  • C. C. The electric field strengths at points A and B are equal
  • D. D. The potential energy of a positive charge at $A$ is greater than that at $B$.

Answer: A

Solution: The potential energy of the positive charge at $A$ is smaller than that at $B$, option $B D$ is wrong; the sparseness of the electric field lines reflects the magnitude of the field strength, so it can be seen that the electric field strength at $A$ point is greater than that at $B$, option $B$ is wrong; option [[]] is wrong. INLINE_FORMULA_8]] is wrong.

Question 23: 24. Using electric field lines makes it very visual and easy to compare the strength of the field at...

24. Using electric field lines makes it very visual and easy to compare the strength of the field at various points in an electric field. Figure A is an equal number of dissimilar point charges to form an electric field of electric field lines, Figure B is some points in the field: $O$ is the midpoint of the line of charge, $a , b$ is the line of the plumb line on the symmetry of the relative $O$ two points, $c , d$ is the line of symmetry of the relative $O$, $c , d$ is the line of symmetry of the relative $O$. _3]] are two points of symmetry on the line with respect to $O$. then ![](/images/questions/phys-electrostatics/image-008.jpg) A ![](/images/questions/phys-electrostatics/image-009.jpg) B

  • A. A. $a , b$ field strength is the same, potential is equal
  • B. C. $c , d$ field strength is the same, potential is equal
  • C. B. $a , b$ field strengths are not equal, potentials are not equal
  • D. D. The ... $d$ field strengths are not the same and the potentials are not equal

Answer: A

Solution: AB. Since $a , b$ is the equal amount of heterogeneous point charges on the line of the plumb line about $O$ symmetry of the two points, from the figure A know, $a , b$ at the same degree of sparsity of the electric field lines, and the two points of the electric field lines of the point of the tangent direction of the line are horizontally to the right, so [[[]] two points of field strength is the same; if a charge is moved along the plumb line, due to the direction of electric field force is perpendicular to the direction of movement, the electric field strength is the same. INLINE_FORMULA_3]] have the same field strength; if a charge is moved along the center vertical line, since the direction of the electric field force is perpendicular to the direction of the motion, the electric field force does not do work, so it can be seen that the potentials at the two points of the $a , b$ are equal, so A is correct and B is wrong; CD . Since $c , d$ is a symmetric point on the line of equal and opposite charges about $O$, the electric potentials at the two points of $c , d$ are equal. The sparseness of the field lines is the same, so $c , d$ two points of the field strength is equal in size, the direction of $c \rightarrow d$, so $c , d$ two points of the field strength is the same; according to the direction of the electric field lines, the electric potential decreases gradually, so it can be seen that the electric potential of the $c$ point is the same. _11]] is higher than that at $d$, so CD is wrong. Therefore, CD is wrong.

Question 24: 25. The following is a correct statement about electric charge

25. The following is a correct statement about electric charge

  • A. A. The meta-charge is the electron
  • B. B. Induction creates an electric charge
  • C. C. The international unit of charge is the coulomb
  • D. D. The Coulomb force between charges is an effect force

Answer: C

Solution: A. Meta-charge refers to the smallest amount of charge, not protons or electrons, so A is wrong; B. Charge can neither be created nor destroyed, but can only be transferred from one object to another, or from one part of an object to another part, the total amount is conserved, so B error; C.In the International System of Units, the unit of charge is the coulomb, so C is correct; D. The Coulomb force between charges is able to produce over the electric field, that is, a field force, so D error.

Question 25: 26. As shown in the figure, the upper end of two insulated wires are tied to the horizontal ceiling,...

26. As shown in the figure, the upper end of two insulated wires are tied to the horizontal ceiling, and the other end is connected to two charged balls. $P , Q$, the balance of the two spheres in the same horizontal line, the two wires and the ceiling between the angle were $\alpha = 30 ^ { \circ } , \beta = 45 ^ { \circ }$, the acceleration of gravity is ${ } ^ { g }$, now cut the thin line connecting the $\mathrm { Q } ^ { \mathrm { T } }$ balls, then in the moment of cutting the thin line, the Q $\mathrm { Q } _ { \text {球的加速度大小为( )} }$ ![](/images/questions/phys-electrostatics/image-010.jpg)

  • A. A. $\sqrt { 2 } g$
  • B. B. $\sqrt { 3 } g$
  • C. C. 2 g
  • D. D. $\sqrt { 6 } g$

Answer: A

Solution: Before the thread is cut, the tension in the thread connecting ball Q is $T = \frac { m g } { \cos 45 ^ { \circ } } = \sqrt { 2 } m g$. After the thread is cut, the downward gravitational force on ball Q and the gravitational force of P on Q remain unchanged, so the combined force is $F _ { \text {合 } } = T = \sqrt { 2 } m g = m a$ solving for $a = \sqrt { 2 } g$.

Question 26: 27. The Taiji diagram, with its rich and complex meaning, embodies the wisdom of ancient Chinese phi...

27. The Taiji diagram, with its rich and complex meaning, embodies the wisdom of ancient Chinese philosophy. As shown in the figure, $O$ is the center of the great circle, $O _ { 1 }$ is the center of the upper yang semicircle, $O _ { 2 }$ is the center of the lower yin semicircle, $O , O _ { 1 } , O _ { 2 }$ is in the same straight line, $A B$ is the diameter of the great circle and is connected to ${ } ^ { O } O _ { 2 }$. FORMULA_4]] is the diameter of the great circle and perpendicular to the ${ } ^ { O } O _ { 2 }$ line, $C , D$ is the two points of symmetry about the $O$ point, and the ${ } ^ { O } { } ^ { 1 } , { } ^ { O _ { 2 } }$ points are fixed to the two points of the same size of the charge. The electric field generated by the point charge at $O _ { 1 } , O _ { 2 }$ is the only electric field in the entire space. The following statement is correct ![](/images/questions/phys-electrostatics/image-011.jpg)

  • A. A. $C , D$ Two points of equal electric potential
  • B. B. The potential energy of an electron increases and then decreases as it is moved from $A$ along a straight line to $B$.
  • C. C. As the proton is moved from $A$ to $B$ in a straight line, the electric field force on the proton first increases and then decreases
  • D. D. An electron (regardless of gravity) is released from rest at $A$, and the electron can move back and forth between $A B$ along a straight line.

Answer: C

Solution: A. Fix dissimilar point charges of equal magnitude at ${ } ^ { O } , { } ^ { O }$, and let ${ } ^ { O }$ be a positive point charge, and ${ } ^ { O }$ be a negative point charge, and since $C$ is close to the positive point charge, and [$D$ is close to the negative point charge, then $C$ has a higher potential than $D$, then $C$ is close to the negative point charge. INLINE_FORMULA_4]] is close to the negative charge, the potential at $C$ is higher than that at $D$, so A is wrong; B. $A B$ is the middle vertical line of the line of equal dissimilar charges, according to the characteristics of the distribution of potential of equal dissimilar charges, the middle vertical line is an equipotential line, so the potential energy of the electrons remains unchanged in the process of moving the electrons from $A$ to $B$ along the straight line, so the potential energy of the electrons remains unchanged, so $B$ is the same. $B$ is wrong; C. According to the equal amount of heterogeneous charges in the plumb line electric field distribution characteristics can be known, $O$ point for the plumb line on the field strength of the largest point, then the mass of the $O$ point INLINE_FORMULA_11]] is the point with the strongest field on the plumbline, then the proton is moved from $A$ to $B$ along a straight line, the field strength becomes bigger and then smaller, and the electric field force on the proton increases and then decreases, so $C$ is correct; D. Since according to the equal amount of dissimilar charges on the center plumb line of the field strength direction perpendicular to the center plumb line, an electron (not counting the weight) from the $C$ will be D. Since an electron (without weight) is released from $A$ at $A$, the electric field force is perpendicular to the line at $A B$ and it is impossible for the electron to move back and forth along the straight line between $A B$, so D is wrong.

Question 27: 28. Thunderstorm days in the vicinity of the lightning rod to produce electric fields, the distribut...

28. Thunderstorm days in the vicinity of the lightning rod to produce electric fields, the distribution of its equipotential surface as shown in the figure in the dotted line shown in the lightning rod directly above the establishment of the horizontal direction of the coordinate axis $O x$, the axis of the distribution of the potential of the various points as shown in the figure, $b$ for the highest point of the graph line, then ( ) ![](/images/questions/phys-electrostatics/image-012.jpg) ![](/images/questions/phys-electrostatics/image-013.jpg)

  • A. A. $b$ point field strength is zero
  • B. C. $a , c$ The field strength is the same at both points
  • C. B. $b$ Point field is strong and straight up.
  • D. D. The $b$ point is further away from the lightning rod than the $c$ point.

Answer: B

Solution: AB. According to the absolute value of the slope of the $\varphi ^ { - x }$ image is the magnitude of the electric field strength, it can be seen that the electric field strength in the horizontal direction of the $b$ point in the figure is zero, but the vertical component of the field strength is not zero, and according to the relationship between the electric field lines and the equipotential surface, it can be seen that the component in the vertical direction is straight up. Therefore, A is wrong and B is correct; C. According to the direction of the electric field lines along the potential decrease can be known, along the $x$ axis positive direction $a$ point near the potential increases, $c$ point near the potential decreases, so the $a , c$ two points in the horizontal direction of the field strength component direction is opposite, and the $a , c$ two points of the field strength component direction is opposite, and the component in the solid direction is solid up. The field components in the horizontal direction of $a , c$ are opposite to each other, and the field components in the straight direction are upward, so the field directions of $a , c$ must be different, and C is wrong; D. As can be seen from the figure, the position of the equipotential surface directly above the lightning rod tangent to the horizontal, where the electric field lines straight up, the $b$ point may be in the lightning rod directly above, so along the horizontal direction, $b$ point from the lightning rod distance is likely to be zero, D error.

Question 28: 29. Two identical electrically charged metal spheres with charges of $+ 3 q$ and $+ 7 q$ have radii ...

29. Two identical electrically charged metal spheres with charges of $+ 3 q$ and $+ 7 q$ have radii that are much smaller than the distance between the centers of the two spheres $r$. When they are placed back in contact, the magnitude of the electrostatic force at this point is ( )

  • A. A. $k \frac { 10 q ^ { 2 } } { r ^ { 2 } }$
  • B. B. $k \frac { 16 q ^ { 2 } } { r ^ { 2 } }$
  • C. C. $k \frac { 21 q ^ { 2 } } { r ^ { 2 } }$
  • D. D. $k \frac { 25 q ^ { 2 } } { r ^ { 2 } }$

Answer: D

Solution: Two identical metal spheres with charges $+ 3 q$ and $+ 7 q$ are brought into contact with $$ \frac { 3 q + 7 q } { 2 } = 5 q $$ By Coulomb's law, the electrostatic force at this point is given by $$ F = k \frac { ( 5 q ) ^ { 2 } } { r ^ { 2 } } = k \frac { 25 q ^ { 2 } } { r ^ { 2 } } $$

Question 29: As shown in the figure, two charged spheres $a , b$ are located on the two vertices of a triangle wi...

As shown in the figure, two charged spheres $a , b$ are located on the two vertices of a triangle with side length $l$, $a , b$ are positively charged, and both have charge $q$. ]]. The electrostatic force constant is known to be $k$. Find the magnitude of the electric field strength at the vertex $c$ to be ![](/images/questions/phys-electrostatics/image-014.jpg)

  • A. A. $\frac { \sqrt { 3 } k q } { l ^ { 2 } }$
  • B. B. $\frac { \sqrt { 3 } k q } { 3 l ^ { 2 } }$
  • C. C. $\frac { 3 k q } { l ^ { 2 } }$
  • D. D. $\frac { 2 \sqrt { 3 } k q } { l ^ { 2 } }$

Answer: A

Solution: From the meaning of the question, $a , b$, the two positive charges at $c$ are equal in magnitude, and the directions are at an angle of 60 degrees to each other, so the combined field strength at $c$ is $$ E ^ { \prime } = 2 E \cos 30 ^ { \circ } = 2 \times \frac { k q } { l ^ { 2 } } \times \frac { \sqrt { 3 } } { 2 } = \frac { \sqrt { 3 } k q } { l ^ { 2 } } $$

Question 30: 31. As shown in the figure, two point charges are fixed at $O , O ^ { \prime }$ two points, along th...

31. As shown in the figure, two point charges are fixed at $O , O ^ { \prime }$ two points, along the $x$ axis direction of the potential with the position coordinate of the change rule of the pattern as shown in the figure, $a , c$ the two points of the same longitudinal coordinates of the value of $b$ point is the lowest point of the image. The $b$ point is the lowest point of the image. It is known that $O O ^ { \prime } : O b = 3 : 2$. The following statements are correct ( ) ![](/images/questions/phys-electrostatics/image-015.jpg)

  • A. A. The point charges at the points $O , O ^ { \prime }$ are equal and dissimilar and the point $O$ is negatively charged
  • B. B. $O , O ^ { \prime }$ The ratio of the absolute values of the charges carried by the point charges at the two points is $4 : 9$
  • C. C. The direction of the electric field between $a C$ is along the positive direction of the $x$ axis.
  • D. D. Electronic from $a _ { \text {点沿 } x \text { 轴正方向移动到 } C \text { 点的过程,电场力先做负功后做正功 } }$

Answer: D

Solution: AC. According to the decreasing potential along the electric field lines, it can be seen that the direction of field strength between $a , b$ is in the positive direction along the $x$ axis, and the direction of field strength between b, ${ } ^ { c }$ is in the negative direction along the ${ } ^ { X }$ axis, so the point at ${ } ^ { \prime }$ is negatively charged, and the point at ${ } ^ { O }$ is negatively charged. INLINE_FORMULA_6]] is negatively charged, and ${ } ^ { O }$ is positively charged, so AC is incorrect; B. The absolute value of the slope of the $\varphi ^ { - x }$ image indicates the magnitude of the field strength, and it can be seen from the figure that the field strength of the $b$ point is zero, and it can be seen from the formula of the field strength of the point charge as well as the principle of superposition of electric field strength. $$ k \frac { q } { ( O b ) ^ { 2 } } = k \frac { q ^ { \prime } } { \left( O ^ { \prime } b \right) ^ { 2 } } $$ Also $O O ^ { \prime } : O b = 3 : 2$, solve for $$ q : q ^ { \prime } = 4 : 25 $$ Therefore, B is wrong; D. The electric field force on the electron in the $a b$ section is along the negative direction of the $x$ axis, then the electron goes from $a$ to $b$ and the electric field force does negative work, and the electrons in the $b c$ section do negative work. FORMULA_15]] is along the $x$ axis, then the electrons from $b$ to $c$ do positive work, so D is correct.

Question 31: 32 . With respect to electric field strength and magnetic induction, the following statements are co...

32 . With respect to electric field strength and magnetic induction, the following statements are correct ( )

  • A. A. From the equation $\mathrm { E } = \mathrm { k } \frac { Q } { r ^ { 2 } }$ for the electric field strength of a point charge in a vacuum, it follows that as r tends to zero, its electric field strength tends to infinity
  • B. B. Definitional equation $\mathrm { E } = \frac { F } { q }$ applies to any electric field.
  • C. C. From the formula for the amperometric force $F = B I L$, it is clear that a small piece of energized conductor is not subject to the amperometric force at a certain place, which means that there must be no magnetic field here
  • D. D. The greater the force on an energized wire in a magnetic field, the stronger the field is

Answer: B

Solution: When r tends to zero, the two charges can not be seen as a point charge, Coulomb's law can not be established, so option A is wrong; the definition of the electric field strength of $E = \frac { F } { q }$ applies to any electric field, option B is correct; a small section of the conductor is not subject to the Ampere force at a certain place, may be the direction of the current is parallel to the direction of the magnetic field, and this is not necessarily the absence of a magnetic field, option C is wrong; by the Ampere By the Ampere force formula $\mathrm { F } = \mathrm { BIL }$ can be known, the energized wire in the magnetic field, the greater the force, the magnetic field is not necessarily the stronger, Option D is wrong. [EYE POINT] This question examines the understanding of the electric field strength, magnetic induction strength of the formula; to know the scope of Coulomb's law, know the application of the three electric field strength formula; when the energized wire is placed in the magnetic field by the Ampere force is related to the placement of the wire, parallel to zero, perpendicular to the maximum.

Question 32: 33. In the International System of Units, the unit of the electrostatic force constant $k$ is ( )

33. In the International System of Units, the unit of the electrostatic force constant $k$ is ( )

  • A. A. $\frac { C ^ { 2 } } { m }$
  • B. B. $\frac { m ^ { 2 } } { C ^ { 2 } }$
  • C. C. $\frac { N \cdot C ^ { 2 } } { m ^ { 2 } }$
  • D. D. $\frac { N \cdot m ^ { 2 } } { C ^ { 2 } }$

Answer: D

Solution: According to Coulomb's law $$ F = k \frac { Q q } { r ^ { 2 } } $$ we have $$ k = \frac { F r ^ { 2 } } { Q q } $$ The units of force $F$, distance $r$, and charge $( Q , q )$ in the International System of Units (SIU) are, respectively, $N , m , C$, and so $k$ is in units of $\mathrm { N } \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$

Question 33: 34. As shown in the figure, in the radius of $r$ on the circle there are four points exactly constit...

34. As shown in the figure, in the radius of $r$ on the circle there are four points exactly constitute a square, in the corresponding point fixed point charge, point charge charge has been marked in the figure, then the center of the circle $O$ at the size of the electric field strength (electrostatic force constant for $k$) ( ) () 2]]) ( ) ![](/images/questions/phys-electrostatics/image-016.jpg)

  • A. A. $\frac { 3 k Q } { r ^ { 2 } }$
  • B. B. $\frac { k Q } { r ^ { 2 } }$
  • C. C. $\frac { 5 k Q } { r ^ { 2 } }$
  • D. D. 0

Answer: B

Solution: According to $E = \frac { k Q } { r ^ { 2 } }$, $- Q$ and -$Q$ at point $O$ generate field strengths of equal magnitude and opposite direction, which cancel each other, and $2 Q$ at point $O$ and $Q$ at point $Q$ at $2 Q$ cancel each other. FORMULA_5]] at $O$ is equal in magnitude and opposite in direction to $Q$ at $O$. $$ E = \frac { 2 k Q } { r ^ { 2 } } - \frac { k Q } { r ^ { 2 } } = \frac { k Q } { r ^ { 2 } } $$

Question 34: 35. As shown in the figure for a part of the electrostatic field electric field line distribution, t...

35. As shown in the figure for a part of the electrostatic field electric field line distribution, the following statements are correct ![](/images/questions/phys-electrostatics/image-017.jpg)

  • A. A. The electric field strength at point $a$ is less than the electric field strength at point $b$.
  • B. B. The potential at point $a$ is lower than that at point $b$.
  • C. C. This electric field could be the one formed by a positive point charge
  • D. D. This electric field may be formed by a negative point charge

Answer: B

Solution:

Question 35: 36. As shown in the figure, the square wire frame consists of insulated rods of uniform thickness wi...

36. As shown in the figure, the square wire frame consists of insulated rods of uniform thickness with side length $L$, and the positive charge is evenly distributed on the wire frame, with the center of the wire frame $O$ as the origin and the establishment of the $x O y$ planar right-angled coordinate system, and now, a sufficiently short piece of wire frame is taken off the center of the upper side of the wire frame parallel to the $x$ axis at $A$. INLINE_FORMULA_3]] parallel to the $x$ axis, now take a short enough piece of $q$ charged with $q$ from the midpoint of the upper side of the wire frame parallel to the $x$ axis, and move it to $B$ in the frame, and move it to $B$ in the $O$ axis. FORMULA_7]] point in the wireframe, if the wireframe its If the charged quantity and charge distribution of the other parts of the wire frame remain unchanged, at this time, the direction of the electric field strength at the $O$ point is positively oriented to the $x$ axis at the $37 ^ { \circ }$ angle upward, and if it is known that the $\sin 37 ^ { \circ } = 0.6 , \cos 37 ^ { \circ } = 0.8$, the $B$ will be the same as the $B$ point. _FORMULA_12]] point has positional coordinates ( ) ![](/images/questions/phys-electrostatics/image-018.jpg)

  • A. A. $\left( - \frac { \sqrt { 2 } } { 3 } L , 0 \right)$
  • B. B. $\left( - \frac { \sqrt { 3 } } { 4 } L , 0 \right)$
  • C. C. $\left( - \frac { \sqrt { 2 } } { 4 } L , 0 \right)$
  • D. D. $\left( - \frac { \sqrt { 3 } } { 6 } L , 0 \right)$

Answer: B

Solution: The combined field strength of the entire square uniformly positively charged wireframe at $O$ is zero, and it can be seen that a sufficiently short charged wireframe will be taken off at $A$, and a short charged wireframe will be taken off at $A$. A short section of the wireframe at $q$ with a charge of $q$ is taken off, and the remaining section at $O$ produces a field of magnitude $+ y$ in the direction of $+ y$. $$ E _ { y } = k \frac { q } { \left( \frac { L } { 2 } \right) ^ { 2 } } = \frac { 4 k q } { L ^ { 2 } } $$ Move a small section with a charge of $q$ taken from $A$ to point $B$ on the $x$ axis, and set the point $B$ away from the $B$ point. $O$ at point $r$. It can be shown that the field strength of this small section at point $O$ is in the direction of $+ x$, and the magnitude is $$ E _ { x } = \frac { k q } { r ^ { 2 } } $$ According to the question $$ \tan 37 ^ { \circ } = \frac { E _ { y } } { E _ { x } } $$ The joint solution is $$ r = \frac { \sqrt { 3 } } { 4 } L $$ The coordinates of $B$ are $\left( - \frac { \sqrt { 3 } } { 4 } L , 0 \right)$, B is correct and ACD is wrong.

Question 36: 37. As shown in the figure, a metal block A fixed on a smooth horizontal surface in an automatic con...

37. As shown in the figure, a metal block A fixed on a smooth horizontal surface in an automatic control instrument is connected to a metal slider B by an insulated spring with an original length of $L$, which carries an equal amount of the same kind of charge, and the spring extension is ${ } ^ { ~ } { } ^ { 1 }$. If the $\mathrm { A } , \mathrm {~B}$ charges are halved, the spring extension becomes ${ } ^ { ~ X _ { 2 } }$ and the slider can be seen as a mass, then ( ) ![](/images/questions/phys-electrostatics/image-019.jpg)

  • A. A. $x _ { 2 } = \frac { 1 } { 2 } x _ { 1 }$
  • B. B. $x _ { 2 } = \frac { 1 } { 4 } x _ { 1 }$
  • C. C. $x _ { 2 } > \frac { 1 } { 4 } x _ { 1 }$
  • D. D. $x _ { 2 } < \frac { 1 } { 4 } x _ { 1 }$

Answer: C

Solution: The Coulomb force is equal to the elastic force, and when the charge of each of the two balls is halved, if we do not take into account the effect of the change in the distance between the two balls on the Coulomb force, by $$ F = k \frac { q _ { 1 } q _ { 2 } } { r ^ { 2 } } $$ Coulomb force can be reduced to the original $\frac { 1 } { 4 }$, the $$ x _ { 2 } = \frac { 1 } { 4 } x _ { 1 } $$ But the reality is that the Coulomb force increases again when the distance decreases, therefore $$ x _ { 2 } > \frac { 1 } { 4 } x _ { 1 } $$

Question 37: 38. As shown in the figure, a point charge of $+ Q$ is fixed at point $+ Q$ of $C$ of a positive tri...

38. As shown in the figure, a point charge of $+ Q$ is fixed at point $+ Q$ of $C$ of a positive triangular prism $A B C - A ^ { \prime } B ^ { \prime } C ^ { \prime }$, and a point charge of $- Q$ at point $D , D ^ { \prime }$ of $D , D ^ { \prime }$, respectively, and the point $D , D ^ { \prime }$ is the point $A C , A ^ { \prime } C ^ { \prime }$. FORMULA_4]], the point $D , D ^ { \prime }$ is the midpoint of the $A C , A ^ { \prime } C ^ { \prime }$ side respectively, and the potential is zero at infinity. The following statements are correct ( ) ![](/images/questions/phys-electrostatics/image-020.jpg)

  • A. A. The electric field strength at points B and B is the same
  • B. B. Moving a negative test charge from point $A$ to point $C$ increases its potential energy
  • C. C. A positive test charge is moved along a straight line from point $B$ to point $D$ and the electric field force does positive work
  • D. D. If a point charge with charges $+ Q$ and $- Q$ is fixed at points $A ^ { \prime }$ and $C$, respectively, then the direction of the field at point $D$ points points towards $B$ and $B$. [INLINE_FORMULA_5]].

Answer: B

Solution: A.The two points of $B , B$ are on the middle vertical line of the $A C$ line, and according to the characteristics of the distribution of the electric field lines of the equal amount of dissimilar charges, it can be seen that B, ${ } ^ { B }$ are both parallel to ${ } ^ { A C }$ and in the same direction, but ${ } ^ { B }$ is farther away from ${ } ^ { A C }$, so the electric field strength at ${ } ^ { B }$ is greater than that at ${ } ^ { A C }$. The electric field at ${ } ^ { B }$ is greater than that at $B$, A is wrong; B. According to the characteristics of the distribution of electric field lines of equal amounts of dissimilar charges, the electric potential at point $A ^ { \prime }$ is higher than that at point $C ^ { \prime }$. Therefore, the negative test charge is moved from point $A ^ { \prime }$ to point $C$, and the force of the electric field does negative work, and the potential energy increases. B is correct; C. From the distribution of the electric field lines of an equal number of dissimilar charges, it can be seen that the surface $B ^ { \prime } D ^ { \prime } D$ is an equipotential surface, and the electric field force does not do any work when a positive test charge is moved along a straight line from the point $B ^ { \text {点 } }$ to the point $D$, and the C is incorrect; D. Fix another point charge of $+ Q$ at point $A ^ { \prime }$ and another point charge of -$Q$ at point $C ^ { \prime }$, and the electric field force will not do any work, C is incorrect; D. Fix another point charge of $A ^ { \prime }$ at point $+ Q$, and then another point charge of -$Q$ at point $Q$. principle of superposition of electric field strength and field strength direction characteristics, it can be seen that the $A , C$ two point charges at the $D$ point combine the field strength direction along the $A C$ direction, and the $A ^ { \prime }$ and $C$ points are along the $C$ and $C$ points, which is a good example of the superposition principle and field strength direction characteristics. FORMULA_23]] and $D$ are in the $D$ direction, so the overall direction of the combined field is toward the $C$ point, D is wrong. D is wrong.

Question 38: 39. Snowflakes hemp oil and hair crumbs placed in the vessel, with the starter to make the electrode...

39. Snowflakes hemp oil and hair crumbs placed in the vessel, with the starter to make the electrode charged, hair crumbs will be presented as shown in the figure, known as the right terminal connected to the positive pole of the starter. Then the following statements are correct ( ) ![](/images/questions/phys-electrostatics/image-021.jpg)

  • A. A. Electric field lines are real.
  • B. B. $a$ point field strength less than $b$ point field strength
  • C. C. The $a$ point potential is higher than the $b$ point potential.
  • D. D. Move a positive test charge from point $a$ to point $b$, the electric field force does negative work

Answer: D

Solution: A. The electric field lines are imaginary, not real, so A is wrong; B. The black lines in the figure describe the distribution of the electric field. The electric field lines at $a$ are denser than those at $b$, and the field at $a$ is stronger than that at $b$, so it is wrong to say that B. The electric field lines are imaginary, not real, so A is wrong; B. The black lines in the figure describe the distribution of the electric field; CD. The potential at point $a$ is lower than that at point $b$, and the electric field force does negative work when a positive test charge is moved from point $a$ to point $b$, so C is incorrect and D is correct.

Question 39: 40. As shown in the figure, a uniformly positively charged solid insulating ball inside a smooth-wal...

40. As shown in the figure, a uniformly positively charged solid insulating ball inside a smooth-walled straight pipe $A C$ (the pipe is located in the horizontal plane, its inner diameter can be ignored), the center of the ball $O$ in the $A C$ on the line, the pipe is not charged inside. A negatively charged sphere (which can be considered as a mass) is released from rest at point $A$ of the pipe and moves towards point $C$ only under the action of the electric field force. It is known that the force of a uniformly charged spherical shell on a charged ball inside it is zero, not counting the gravitational force between the two balls, then the following statements are correct ( ) ![](/images/questions/phys-electrostatics/image-022.jpg) High School Physics Assignment, October 29, 2025

  • A. A. $A \rightarrow O$ process, the ball's potential energy gradually increases
  • B. B. $A \rightarrow O$ process, the ball's acceleration remains constant
  • C. C. The ball will keep accelerating away from the $C$ point and leave the pipe.
  • D. D. The ball has maximum kinetic energy at $O$.

Answer: D

Solution: C. From the analysis, we know that the ball will move in simple harmonic motion between $A , C$, $O$ is the equilibrium position, and $A , C$ is the amplitude of the two points. Therefore, the ball will not leave the pipe at $C$, and option $C$ is wrong; A. $A \rightarrow O$ process, the electric field force has been on the ball to do positive work, the ball's kinetic energy gradually increased, the potential energy gradually decreased, option A is wrong; D. $O$ point is the equilibrium position, the maximum speed, the maximum kinetic energy, option D is correct; B. Let the charge of the insulated ball be $Q$, the radius be $R$, and the charge of the ball be $q$, and the distance of the ball to the point $O$ at a certain moment be ${ } _ { r }$, and the distance of the ball to the point ${ } _ { r }$ is $F = k \frac { \frac { r ^ { 3 } } { R ^ { 3 } } Q q } { r ^ { 2 } } = \frac { k r Q q } { R ^ { 3 } }$, which is the same as that of the insulated ball. FORMULA_11]] , then by Coulomb's law we have $F = k \frac { \frac { r ^ { 3 } } { R ^ { 3 } } Q q } { r ^ { 2 } } = \frac { k r Q q } { R ^ { 3 } }$ Therefore, $A \rightarrow O$ process, the ball by the electric field force has been reduced, combined with Newton's second law can be known as the acceleration has been reduced, option B error.
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Electrostatics

静电场

39 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Electrostatic Fields The study of electric fields generated by stationary charges and their properties is an important foundation module for the CSCA Physics exam. The section focuses on the understanding and application of core concepts such as electric field strength, potential and Coulomb's law. Questions often incorporate scenarios such as electrostatic equilibrium of conductors and superposition of electric fields, requiring strong analytical and computational skills.

Questions:39

Key Points

  • 1Coulomb's law and calculation of electric field force
  • 2Concept and relationship between electric field strength and electric potential
  • 3Properties and applications of electrostatic equilibrium of conductors
  • 4Principle of electric field superposition and typical field distribution

Study Tips

It is recommended to practice by comparing the physical significance and calculation of electric field strength (vector) and potential (scalar) with typical electric field models.

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