In the figure, the inner (shaded) region A re | vedclass

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Question

$q _1=\int_0^1 k r 4 \pi r^2 d c=\frac{4 \pi k}{4}=\pi k$

$q_2=\int_1^2 \frac{2 k}{r} 4 \pi r^2 d r=\frac{8 \pi k\left(r^2-1^2\right)}{2}$

$q_2=

Vedclass · Accepted Answer

If $r_B=\frac{3}{2}$, then the electric potential just outside $B$ is $\frac{k}{\epsilon_0}$.

Vedclass · Answer

$q _1=\int_0^1 k r 4 \pi r^2 d c=\frac{4 \pi k}{4}=\pi k$

$q_2=\int_1^2 \frac{2 k}{r} 4 \pi r^2 d r=\frac{8 \pi k\left(r^2-1^2ight)}{2}$

$q_2=4 \pi k\left[r^2-1ight]=4 \pi k r^2-4 \pi k$

$q_{	ext {net }}=q_1+q_2$

$=4 \pi r^2-3 \pi k$

$q_{p e t}=\pi k\left[4 r^2-3ight]$

$(A)$ $E_{2 x t}=0 \Rightarrow q_{\infty}=0 \Rightarrow r=\frac{\sqrt{3}}{2}$

$(B)$ $V =\frac{ kQ Q _{ R }}{ r }=\frac{1}{4 \pi \varepsilon_0} \frac{\pi k\left(4 r^2-3ight)}{r}$

$V =\frac{ k }{4 \varepsilon_0}\left[4 r-\frac{3}{r}ight]$

$=\frac{k}{4 \varepsilon_0}\left[4 	imes \frac{3}{2}-\frac{3 	imes 2}{3}ight]=\frac{k}{\varepsilon_0}$

$	ext { (C) } q _{2 e }=\pi k \left[4(2)^2-3ight]$

$=13 \pi k$

$(C)$

$q _{2 m}=\pi k \left[4(2)^2-3ight]$

$=13 \pi k$

$(D)$

$E_z =\frac{k Q}{r^2}$

$=\frac{1}{4 \pi \varepsilon_0} \frac{\pi k\left(4 r^2-3ight)}{r^2}$

$=\frac{k}{4 \varepsilon_0}\left[\frac{4\left(\frac{5}{2}ight)^2-3}{(5 / 2)^2}ight]$

$=\frac{k}{25 \varepsilon_0}[25-3]=\frac{22}{25} \frac{k}{\varepsilon_0}$

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