\( E \) and \( V \), respectively, denote electric field intensity and electrostatic potential at a point. The correct relation between the two is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( E = -dV/dr \)
The work done by the field on a unit positive charge moved through \( dr \) equals the decrease in potential: \( E\,dr = -dV \). So the electric field is the negative of the potential gradient.
\( \displaystyle E = -\frac{dV}{dr} \)
The minus sign shows that the field points in the direction in which the potential decreases fastest.
NCERT (2026-27):
Relation Between Field and Potential – “… The work done in this process is \( |\mathbf{E}|\,\delta l \). This work equals the potential difference \( V_A - V_B \). Thus,
\( \displaystyle |\mathbf{E}|\,\delta l = V - (V + \delta V) = -\delta V \)
i.e.,
\( \displaystyle |\mathbf{E}| = -\frac{\delta V}{\delta l} \qquad (2.20) \)”
Relation Between Field and Potential – “… We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.”