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Two isolated, concentric, conducting spherical shells have radii $R$ and $2 R$ and uniform charges $q$ and $2 q$ respectively. If $V_1$ and $V_2$ are potentials at points located at distances $3 R$ and $\frac{R}{2}$, respectively, from the centre of shells.
Then the ratio of $\left(\frac{V_2}{V_1}\right)$ will be
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Then the ratio of $\left(\frac{V_2}{V_1}\right)$ will be
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Verified Answer
The correct answer is:
2
For point, $r=\frac{R}{2}$
$V_1=$ Potential of surface radius $R+$ Potential of surface radius $2 R$
$$
=\frac{k q}{R}+\frac{k 2 q}{2 R}=\frac{2 k q}{R}
$$
For point $\quad r=3 R$,
$V_2=$ Potential due to charges $q$ and $2 q$ assumed to be concentrated at centre $=\frac{k(3 q)}{3 R}=\frac{k q}{R}$
So, ratio $\quad \frac{V_1}{V_2}=2$
$V_1=$ Potential of surface radius $R+$ Potential of surface radius $2 R$
$$
=\frac{k q}{R}+\frac{k 2 q}{2 R}=\frac{2 k q}{R}
$$
For point $\quad r=3 R$,
$V_2=$ Potential due to charges $q$ and $2 q$ assumed to be concentrated at centre $=\frac{k(3 q)}{3 R}=\frac{k q}{R}$
So, ratio $\quad \frac{V_1}{V_2}=2$
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