A charge of \(+2\ \mu\mathrm{C}\) placed at a point P feels a force of \(6\times10^{-3}\ \mathrm{N}\) towards the east. It is then replaced by a charge of \(-4\ \mu\mathrm{C}\). The electric field at P and the force on the new charge are, respectively,
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3\times10^{3}\ \mathrm{N\,C^{-1}}\) east, \(1.2\times10^{-2}\ \mathrm{N}\) west
The electric field at P is set up by the source charges. It does not change when the test charge is changed. So first find \( E \) from the first charge, then use \( F=qE \) for the new charge.
\( \begin{aligned} E &= \frac{F}{q} \\ &= \frac{6\times10^{-3}}{2\times10^{-6}} \\ &= 3\times10^{3}\ \mathrm{N\,C^{-1}} \end{aligned} \)
The force on the positive charge is towards the east, so the field at P points east. It stays east after the charge is changed.
\( \begin{aligned} F^{\prime} &= |q^{\prime}|E \\ &= 4\times10^{-6}\times3\times10^{3} \\ &= 1.2\times10^{-2}\ \mathrm{N} \end{aligned} \)
A negative charge is pushed opposite to the field. Therefore, the force on the new charge is towards the west.
NCERT (2026-27):
Electric Field – “(ii) Note that the electric field E due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q.”
Example 1.7 – “… the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field.”