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 (integer)
12
Solution
<p>$V = 2a{r^2} + b$</p>
<p>$\Rightarrow E = - {{dV} \over {dr}} = - 4ar$</p>
<p>$\Rightarrow {1 \over {4\pi \varepsilon }}{Q \over {{r^2}}} = - 4ar$</p>
<p>$$ \Rightarrow {Q \over {{4 \over 3}\pi {r^3}}} = 3 \times \varepsilon \times ( - 4a) = - 12a\varepsilon $$</p>
<p>$\Rightarrow \lambda = 12$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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