Electrostatics MCQs for NEET — Physics Questions with Answers

Using the formula for potential energy of two charges $U = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}}$. Here $q_1 = 7 \times 10^{-6} C$, $q_2 = -2 \times 10^{-6} C$, and $r_{12} = 9 cm - (-9 cm) = 18 cm = 0.18 m$. $U = (9 \times 10^9) \frac{(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = 9 \times 10^9 \times \frac{-14 \times 10^{-12}}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 J$. This matches the Example 2.5(a) in the NCERT.

The NCERT states for a system of charges in an external field, the total potential energy includes the mutual interaction energy of the charges plus the energy of interaction of each charge with the external electric field. For two charges, this additional part is $q_1V(r_1) + q_2V(r_2)$. (Example 2.5(c) and equation 2.29).

By definition, 'V at a point P is the work done in bringing a unit positive charge from infinity to the point P.' (Section 2.8.1). Also confirmed earlier in Section 2.3, 'Potential at P due to the charge Q is $V(r) = \frac{Q}{4\pi\epsilon_0 r}$'.

The NCERT states, 'The path-independence of work done by an electrostatic field can be proved using the Coulomb’s law... The concept of the potential energy would not be meaningful if the work depended on the path.' (Section 2.2(i)). This highlights the conservative nature of electrostatic forces.

The provided text explains: 'The amount of work done by the external torque will be given by $pE (\cos\theta_0 - \cos\theta_1)$. This work is stored as the potential energy of the system.' (Section 2.8.3).

In the process of assembling a system of charges, when the first charge ($q_1$) is brought from infinity, there are no other charges yet present in the field to exert any electrostatic force on it. Hence, no work is done against an existing electrostatic field. The NCERT implies this when discussing 'Potential energy of a system of charges' (Section 2.7): 'To bring $q_1$ first from infinity to $r_1$, no work is required.'

According to the NCERT text, 'the potential energy of a system of two charges $q_1$ and $q_2$ is $U = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}}$' (Eq. 2.22), which shows direct proportionality to $q_1q_2$ and inverse proportionality to $r_{12}$. Also, 'the potential energy U would be the same... because of path-independence of work for electrostatic force.' Therefore, both (b) and (c) are correct.

The NCERT text states, 'If $q_1q_2 > 0$, potential energy is positive. This is as expected, since for like charges ($q_1q_2 > 0$), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart.' Both positive-positive and negative-negative charges result in a positive product, hence positive potential energy.

The NCERT states, 'For unlike charges ($q_1q_2 < 0$), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative.' This indicates an attractive force.

The NCERT text explains, 'To bring $q_1$ first from infinity to $r_1$, no work is required.' This is because there are no other charges present at that moment to exert a force against which work needs to be done.