NEET 2020 Physics Question 152 Solution

The electric field vector in electrostatics is equal to the negative spatial gradient of the electric potential scalar function: $$\vec{E} = -\vec{\nabla} V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$$ Along any spatial direction $$r$$: $$E = -\frac{dV}{dr}$$

The problem explicitly states that the potential is found to be $$5\text{ V}$$ throughout the entire volume of $$0.2\text{ m}^3$$. This means $$V(x, y, z) = \text{constant} = 5\text{ V}$$. The derivative of any constant with respect to space coordinates is identically zero: $$\frac{\partial V}{\partial x} = 0, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0$$

$$E = \sqrt{0^2 + 0^2 + 0^2} = 0\text{ N C}^{-1}$$ Thus, the magnitude of the electric field in this entire region is zero.

This is identical to the interior of a charged conductor in electrostatic equilibrium: the potential is uniform and constant throughout the interior, and the electric field inside is strictly zero.