What is the mathematical expression for potential gradient?

The correct answer is Option (3) → \( -\displaystyle\frac{dV}{dr} \)

The potential gradient is the rate at which the electric potential \( V \) changes with distance \( r \). It is tied to the electric field by

    \( \displaystyle E = -\frac{dV}{dr} \)

The minus sign shows that the field points in the direction in which the potential falls. So the expression asked for is the derivative of the potential with respect to distance, written here with this minus sign.

Option (1) \( \displaystyle\frac{dv}{rdr} \) — Incorrect. The extra \( r \) in the denominator gives a unit of \( \mathrm{V\,m^{-2}} \), so it is not a rate of change of potential with distance.

Option (2) \( -\displaystyle\frac{dE}{dr} \) — Incorrect. This is the rate of change of the electric field, not of the potential.

Option (3) \( -\displaystyle\frac{dV}{dr} \) — Correct. It is the derivative of the potential \( V \) with respect to the distance \( r \), with the sign that links it to the field.

Option (4) \( -\displaystyle\frac{dV}{dA} \) — Incorrect. A gradient is taken with respect to distance, not area; \( dV/dA \) has the unit \( \mathrm{V\,m^{-2}} \).

NCERT (2026-27):
Relation Between Field and Potential – “… The work done in this process is |E|δl. This work equals the potential difference V_A – V_B. Thus,

    \( \displaystyle |E|\,\delta l = V - (V + \delta V) = -\,\delta V \quad \text{i.e.,} \quad |E| = -\frac{\delta V}{\delta l} \qquad (2.20) \)

… 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.”