The electric field is defined as \( \vec{E} = \lim\limits_{q_0 \rightarrow 0} \displaystyle\frac{\vec{F}}{q_0} \) where \( q_0 \) is the test charge and \( \vec{F} \) force on \( q_0 \) due to the source charge. The meaning of \( \lim\limits_{q_0 \rightarrow 0} \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The source charge must remain at its original location and its electric field is not affected
A test charge also exerts a force on the source charge. If it were large, it could push the source charges out of their places and change the very field it is meant to measure. Taking the limit \( q_0 \rightarrow 0 \) makes the test charge so small that its effect on the source charge is negligible: the source stays where it is and its field is the one measured.
Option (1) The source charge must remain at its original location and its electric field is not affected — Correct. This is the purpose of the limit: the source is not disturbed, so its field is unchanged.
NCERT (2026-27):
Electric Field – “… Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F is then negligibly small but the ratio F/q is finite and defines the electric field:
\( \displaystyle \mathbf{E} = \lim_{q \to 0} \left( \frac{\mathbf{F}}{q} \right) \qquad (1.9) \)”