Question 649: Electrostatics & Capacitance

Question

For a charged spherical ball, electrostatic potential inside the ball varies with $$r$$ as $$\mathrm{V}=2ar^2+b$$.
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 the medium

Accepted Answer

$$V = 2a{r^2} + b$$
$$ \Rightarrow E =  - {{dV} \over {dr}} =  - 4ar$$
$$ \Rightarrow {1 \over {4\pi \varepsilon }}{Q \over {{r^2}}} =  - 4ar$$
$$ \Rightarrow {Q \over {{4 \over 3}\pi {r^3}}} = 3 	imes \varepsilon  	imes ( - 4a) =  - 12a\varepsilon $$
$$ \Rightarrow \lambda  = 12$$

Electrostatics & Capacitance

Integer / Numeric Answer Type