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  • For a charged spherical ball, electrostatic potential inside the ball varies with as <img alt="\mathrm{V}=2 a r^2+b" src="htt

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

Option: 1

12


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

$$ E=-\frac{d v}{d r}=-4 a r\\

By the Gauss’ theorem

\begin{aligned} & \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{q_{\text {inside }}}{\varepsilon} \\ & E \times 4 \pi r^2=\frac{\rho \times \frac{4}{3} \pi r^3}{\varepsilon} \\ & E=\frac{\rho r}{3 \varepsilon}=-4 a r \\ & \rho=-12 a \varepsilon \end{aligned}

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