Obj-2 Electric Field and Potential HC Verma Solutions Ch-29 Class-12 Vol-2 Concept of Physics for Class-12. Step by Step Solutions of Objective -2 (MCQ-2) Questions of Chapter-29 Electric Field and Potential  (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-12 Physics.

## Obj-2 Electric Field and Potential  (MCQ-2) HC Verma Solutions Ch-29 Vol-2 Concept of Physics for Class-12

 Board ISC and other board Publications Bharti Bhawan Publishers Chapter-29 Electric Field and Potential Class 12 Vol 2nd writer HC Verma Book Name Concept of Physics Topics Solution of Objective-2 (MCQ-2) Questions Page-Number 120

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Exercise

### Obj-2 Electric Field and Potential (MCQ-2)

HC Verma Solutions of Ch-29  Vol-2 Concept of Physics for Class-12

(Page-120)

#### Question-1 :-

Mark out the correct options.

(a) The total charge of the universe is constant.

(b) The total positive charge of the universe is constant.

(c) The total negative charge of the universe is constant.

(d) The total number of charged particles in the universe is constant.

The option (a) The total charge of the universe is constant. is correct

Explanation:

According to the principal of conservation of charge, the net amount of positive charge minus the net amount of negative charge in the universe is always constant. Thus, the total charge of the universe is constant. The total positive charge of the universe may increase or decrease, depending on the total increase or decrease in negative charge. This is the principle of conservation of charge that is universal in nature.

#### Question-3 :-

The electric field and the electric potential at a point are E and V, respectively.

(a) If E = 0, V must be zero.

(b) If V = 0, E must be zero.

(c) If E ≠ 0, V cannot be zero.

(d) If V ≠0, E cannot be zero.

(e) None of the above.

The option (e) None of the above. is correct

Explanation:

Electric field, E=−dV/dr where V = electric potential
For E = 0,  V should be constant.
So, when E = 0,  it is not necessary that V should be 0.
Hence, none of the above signifies the correct relation.

#### Question-4 :-

Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin

(a) must be equal to 20 Vcm−1
(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1
(d) may be less than 20 Vcm−1

The options (b) and (c). are correct

Explanation:

Change in the electric potential, dV = 40 V
Change in length, ∆r = −1−1 = −2 cm

Electric field, E=−dV/dr

⇒ E =−40V/−2

⇒ E = 20Vcm−1

This is the value of the electric field along the x axis.
Electric field is maximum along the direction in which the potential decreases at the maximum rate. But here, direction in which the potential decreases at the maximum rate may or may not be along the x-axis. From the given information,the direction of maximum decrease in potential cannot be found out accurately. So, E can be greater than 20 V/cm in the direction of maximum decrease in potential.
So, the electric field at the origin may be equal to or greater than 20 Vcm−1.

#### Question-5 :-

Which of the following quantities does not depend on the choice of zero potential or zero potential energy?
(a) Potential at a point
(b) Potential difference between two points
(c) Potential energy of a two-charge system
(d) Change in potential energy of a two-charge system

The options (b) and (d). are correct

Explanation:

Potential and potential energy depend on the choice of a reference point of zero potential or zero potential energy. But the difference of potential and energy does not depend on the choice of the reference point. Hence, the correct options are (b) and (d).

#### Question-7 :-

A proton and an electron are placed in a uniform electric field.

(a) The electric forces acting on them will be equal.

(b)  The magnitudes of the forces will be equal.

(c) Their accelerations will be equal.

(d) The magnitudes of their accelerations will be equal.

The option (b)  The magnitudes of the forces will be equal. is correct

Explanation:

We know: F‾= qE‾ For an electron and a proton, the value of q will be same, but the sign will be opposite.
Hence, they will experience a force that will be equal in magnitude but opposite in direction.
Now, As the electron and proton have different values of mass m, they will have different magnitudes of acceleration. Also, they will differ in direction due to the opposite signs of q.

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