JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 25)
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
Answer
12
Explanation
$$V = 2a{r^2} + b$$
$$ \Rightarrow E = - {{dV} \over {dr}} = - 4ar$$
$$ \Rightarrow {1 \over {4\pi \varepsilon }}{Q \over {{r^2}}} = - 4ar$$
$$ \Rightarrow {Q \over {{4 \over 3}\pi {r^3}}} = 3 \times \varepsilon \times ( - 4a) = - 12a\varepsilon $$
$$ \Rightarrow \lambda = 12$$
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