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Chapter 13: Problem 109

Choose the correct statements from the following: (A) If the electric field is zero at a point, the electric potential must also be zero at that point. (B) If electric potential is constant in a given region of space, the electric field must be zero in that region. (C) Two different equipotential surfaces can never intersect. (D) Electrons move from a region of lower potential to a region of higher potential.

Short Answer

Expert verified
The correct statements are (B) "If electric potential is constant in a given region of space, the electric field must be zero in that region" and (C) "Two different equipotential surfaces can never intersect."

Step by step solution

01

Statement A

"If the electric field is zero at a point, the electric potential must also be zero at that point." This statement is not always true. The electric potential at a point depends on the reference point from where we measure the potential. If the electric field is zero at a point, it means that there is no net force acting on a charged particle at that point, but it does not imply that the electric potential is zero.
02

Statement B

"If electric potential is constant in a given region of space, the electric field must be zero in that region." This statement is true. The electric field at a point is the negative gradient of the electric potential with respect to position. If the electric potential is constant within a given region of space, that means the gradient of the potential is zero, which implies that the electric field is zero.
03

Statement C

"Two different equipotential surfaces can never intersect." This statement is also true. Equipotential surfaces are surfaces on which all points have the same electric potential. If two equipotential surfaces were to intersect, it would mean that there would be some points on the intersection with different electric potentials. This contradicts the definition of equipotential surfaces, and hence, two different equipotential surfaces cannot intersect.
04

Statement D

"Electrons move from a region of lower potential to a region of higher potential." This statement is false. Electrons, being negatively charged particles, tend to move from regions of higher potential to regions of lower potential. Hence, the correct statement should be "Electrons move from a region of higher potential to a region of lower potential." Therefore, the correct statements are (B) and (C).

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Most popular questions from this chapter

A charged particle \(q\) is shot towards another charged particle \(Q\) which is fixed, with a speed \(v\). It approaches \(Q\) up to a closest distance \(\mathrm{r}\) and then returns. If \(q\) was given a speed \(2 v\), the closest distance of approach would be (A) \(r\) (B) \(2 r\) (C) \(r / 2\) (D) \(r / 4\)

A conducting hollow sphere of radius \(0.1 \mathrm{~m}\) is given a charge of \(10 \mu \mathrm{C}\). The electric potential on the surface of sphere will be (A) Zero (B) \(3 \times 10^{5} \mathrm{~V}\) (C) \(9 \times 10^{5} \mathrm{~V}\) (D) \(9 \times 10^{9} \mathrm{~V}\)

There are two charges \(+1 \mu \mathrm{C}\) and \(+5 \mu \mathrm{C}\). The ratio of the forces acting on them will be (A) \(1: 5\) (B) \(1: 1\) (C) \(5: 1\) (D) \(1: 25\)

A charge \(q_{1}\) is placed at the centre of a spherical conducting shell of radius \(R\). Conducting shell has a total charge \(q_{2} .\) Electrostatic potential energy of the system (A) \(\frac{q_{1}^{2}+2 q_{1} q_{2}}{8 \pi \varepsilon_{0} R}\) (B) \(\frac{q_{2}^{2}+2 q_{1} q_{2}}{8 \pi \varepsilon_{0} R}\) (C) \(\frac{q_{1}^{2}+q_{1} q_{2}}{4 \pi \varepsilon_{0} R}\) (D) \(\frac{q_{2}^{2}+q_{1} q_{2}}{4 \pi \varepsilon_{0} R}\)

Two capacitors of capacitance \(3 \mu \mathrm{F}\) and \(6 \mu \mathrm{F}\) are charged to a potential of \(12 \mathrm{~V}\) each. They are now connected to each other, with the positive plate of one to the negative plate of the other. Then (A) the potential difference across \(3 \mu \mathrm{F}\) is zero. (B) the potential difference across \(3 \mu \mathrm{F}\) is \(4 \mathrm{~V}\). (C) the charge on \(3 \mu \mathrm{F}\) is zero. (D) the charge on \(3 \mu \mathrm{F}\) is \(10 \mu \mathrm{C}\).
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