For a charged spherical ball, electrostatic potential inside the ball varies with $$r$$ as $$V = 2ar^2 + b$$. Here, $$a$$ and $$b$$ are constant and $$r$$ is the distance from the center. The volume charge density inside the ball is $$-\lambda a\varepsilon$$. The value of $$\lambda$$ is ______.
$$\varepsilon$$ = permittivity of medium.

Solution :

Given electrostatic potential inside the spherical ball :

$$V = 2ar^2 + b$$

Electric field is related to potential by :

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

Therefore,

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

$$E = -4ar$$

Using Gauss law in differential form :

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon}$$

For spherical symmetry :

$$\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}(r^2E_r)$$

Substituting,

$$E_r = -4ar$$

$$\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}(r^2(-4ar))$$

$$= \frac{1}{r^2}\frac{d}{dr}(-4ar^3)$$

$$= \frac{1}{r^2}(-12ar^2)$$

$$= -12a$$

Hence,

$$\frac{\rho}{\epsilon} = -12a$$

Therefore,

$$\rho = -12a\epsilon$$

Comparing with :

$$-\lambda a\epsilon$$

we get,

$$\lambda = 12$$

Final Answer :

$$12$$