JEE Physics 2026 Question with Solution

Given: $V = ar^{3} + b$ inside a spherical region of radius $r$.

Find: The value of $a$ when the total charge in the unit sphere is $a \times \pi \varepsilon_{0}$.

Electric field is related to potential by

$$E = -\frac{dV}{dr}$$

Given

$$V = ar^{3} + b$$

So,

$$E = -\frac{d}{dr}(ar^{3} + b) = -3ar^{2}$$

Using Gauss's law for a spherical surface of radius $r$,

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

Thus,

$$E \cdot 4\pi r^{2} = \frac{Q(r)}{\varepsilon_0}$$

Substituting $E = -3ar^{2}$,

$$(-3ar^{2})(4\pi r^{2}) = \frac{Q(r)}{\varepsilon_0}$$

Hence,

$$Q(r) = -12\pi a \varepsilon_0 r^{4}$$

For the unit sphere, $r = 1$, so

$$Q = -12\pi a \varepsilon_0$$

But given total charge is

$$Q = a \times \pi \varepsilon_{0}$$

Comparing,

$$-12a = a$$

Therefore, the value of $a$ is $-12$.

The correct option is B.

Given: The potential depends only on radial distance, so the field is spherically symmetric.

Find: The constant $a$.

1. Differentiate the potential to get the radial electric field:

$$E(r) = -\frac{dV}{dr} = -3ar^{2}$$

2. For spherical symmetry, Gauss's law gives enclosed charge as

$$Q(r) = \varepsilon_0 E(r) \cdot 4\pi r^{2}$$

3. Substitute the field expression:

$$Q(r) = \varepsilon_0 (-3ar^{2}) \cdot 4\pi r^{2}$$ $$Q(r) = -12\pi a \varepsilon_0 r^{4}$$

4. At $r = 1$,

$$Q(1) = -12\pi a \varepsilon_0$$

5. The question states that the total charge in the unit sphere is $a \pi \varepsilon_0$. Therefore,

$$-12\pi a \varepsilon_0 = a \pi \varepsilon_0$$

This gives the result stated in the solution:

$$a = -12$$

So the correct option is B.