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A spherical capacitor consists of two concentric spherical conductors. The inner one of radius $R_1$ maintained at potential $V_1$ and the outer conductor of radius $R_2$ at potential $V_2$ . The potential at a point P at a distance x from the centre (where $R_2> x> R_1$ ) is

Option 1)
$\frac{V_1-V_2}{R_2-R_1}(x-R_1)$

Option 2)
$\frac{(V_1R_1(R_2-x)+V_2(x-R_1))}{(R_2-R_1)x}$

Option 3)
$V_1+\frac{V_2x}{(R_2-R_1)}$

Option 4)
$\frac{(V_1+V_2)}{R_1+R_2}x$

Answers (1)

best_answer

As we have learned

Potential Difference -

$V= \frac{Q}{4\pi \epsilon _{0}a}-\frac{Q}{4\pi \epsilon _{0}b}$

-

Let $Q_1$ and $Q_2$ be the charges on the inner and the outer sphere respectively. Now is the total potential on the sphere of radius R₁,

So, $V_1 = \frac{Q_1}{R_1} + \frac{Q_2}{R_2}........(1)$

and $V_2$ is the total potential on the surface of sphere of radius $R_2$ ,

So, $V_2 = \frac{Q_2}{R_2} + \frac{Q_1}{R_1}........(2)$

If V be the potential at point P which lies at a distance x from the common centre then

$V = \frac{Q_1}{x} = \frac{Q_1}{R_1} + V_1- \frac{Q_1}{R_1}= Q_1 \left ( \frac{1}{x} - \frac{1}{R_1} \right )+ V_1= \frac{Q_1(R_1-x)}{xR_1}+ V_1$ ……..(iii)

Substracting (ii) from (i)

$V_1- V_2 = \frac{Q_1}{R_1}-\frac{Q_2}{R_2}\Rightarrow (V_1-V_2) R_1R_2= R_2Q_1\Rightarrow Q_1 = \frac{(V_1-V_2)R_1R_2}{R_2-R_1}$

Option 1)

$\frac{V_1-V_2}{R_2-R_1}(x-R_1)$

Option 2)

$\frac{(V_1R_1(R_2-x)+V_2(x-R_1))}{(R_2-R_1)x}$

Option 3)

$V_1+\frac{V_2x}{(R_2-R_1)}$

Option 4)

$\frac{(V_1+V_2)}{R_1+R_2}x$

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Avinash

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