**Q.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.

## Answer

(a) The sum of all charges with the proper sign is always constant in the universe. It can be understood with an example of a neutral atom. The charge is zero. if an electron jumps over to another atom/molecule it will get one unit of positive charge while the receiving atom will get one negative charge, thus the total charge of these two atoms will be still zero.**Q.2. A point charge is brought in an electric field. The electric field at a nearby point**

(a) will increase if the charge is positive

(b) will decrease if the charge is negative

(c) may increase if the charge is positive

(d) may decrease if the charge is negative.

## Answer

(c), (d).Explanation: The field around a nearby point charge will not be the same when it is placed in an electric field whether it is positive or negative. It may increase or decrease depending on the position of the selection of near by point because the electric field will be the vector summation of the fields of the point charge and the original field.

**Q.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.

## Answer

None.Explanation: E is a vector and V is a scaler. At a point, E may be zero in a direction but it may not be zero in the other direction. The direction of E is not clearly stated. Only at a very very far away from a charge E may be zero in all directions and V will be zero here. So it is not necessary that V =0 if E =0 and vice versa(Option (a) and (b) not true). Also, E is zero along a direction where V is constant (option (d) is wrong).

We know that potential V at a point r (vector) where the electric field is E, is given as V = ∫E.dr

Thus even if E is not zero but the angle between E and r is 90°, V will be zero. So option (c) is not true.

**Q.4. The 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}.

## Answer

(b), (c).Explanation: We know,

-(dV/dr) = E.Cosθ

→E = -(1/cosθ)(dV/dr)

=-(1/cosθ)*(-40/2)

=20/Cosθ

Since the value of cosine is from 0 to 1 the value of E at the origin may be from 20 and above depending upon the angle between the x-axis and the direction of vector E. The direction of E is along the maximum rate of change of V.

**Q.5. Which of the following quantities do 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.

## Answer

(b), (d).Explanation: The potential at a point and the potential energy of a two-particle system is defined with respect to a reference point which is the zero potential or zero potential energy. The difference in potential or change in potential energy does not depend upon it because it is with respect to each other.

**Q.6. An electric dipole is placed in an electric field generated by a point charge.**

(a) The net electric force on the dipole must be zero.

(b) The net electric force on the dipole may be zero.

(c) The torque on the dipole due to the field must be zero.

(d) The torque on the dipole due to the field may be zero.

## Answer

(d)Explanation: The arm of a dipole is assumed to be much much less than other dimensions hence the field around it may be assumed to be parallel. Thus there will be equal and opposite electric forces at its two ends and the net force on the dipole will be zero. Since the forces are equal and opposite there will be a torque acting on the dipole in this electric field the magnitude of which will depend on the orientation of the dipole. If the length of the dipole is along the electric field the torque will be zero due to zero arm length of the couple of electric forces.

**Q.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.

## Answer

(b)Explanation: The electric force F on a charge q placed in an electric field E is given as, F = qE,

Since the magnitude of q is equal on a proton and on an electron but it is positive on the proton and negative on the electron so the magnitude of F will be the same on both of them but the direction will be opposite. Hence the option (b) is correct.

The acceleration is given as

a = F/m.

Mass m of an electron is much less than a proton hence the magnitude will not be equal also the direction will be opposite.

**Q.8. The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,**

(a) it is uniform in the region

(b) it is proportional to r

(c) it is proportional to r^{2}

(d) it increases as one goes away from the origin.

## Answer

(c)Explanation: Let the Electric field E = kr. Where r is a constant. Since the difference in the potential is given as

Vr – Vo = ∫Edr, the limit of integration is from origin to distance r. And given Vo =0. Thus,

Vr = ∫krdr =kr²/2

so, Vr ∝ r². The option (c) is correct.