Capacitance of an isolated conducting sphere of radius $R_1$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_2$ connected to earth The ratio of their radii $\left(\frac{ R _2}{ R _1}\right)$ is:

Question

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\dfrac{n}{2}$. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is ___.

Answer 1

To find the ratio of the radii of the spheres, we need to determine the change in the capacitance of the isolated conducting sphere when it is enclosed by a concentric conducting sphere connected to the earth.

The capacitance $ C $ of an isolated conducting sphere of radius $ R_1 $ is given by:

$C = 4 \pi \epsilon_0 R_1$

When the sphere is enclosed by a larger concentric conducting sphere of radius $ R_2 $ connected to earth, the capacitance becomes:

$C' = 4 \pi \epsilon_0 \frac{R_1 R_2}{R_2 - R_1}$

According to the problem, the capacitance increases by a factor of $ n $. Therefore:

$C' = nC$

Substituting the expression for $ C $ and equating, we have:

$4 \pi \epsilon_0 \frac{R_1 R_2}{R_2 - R_1} = n \times 4 \pi \epsilon_0 R_1$

Simplifying this equation gives:

$\frac{R_1 R_2}{R_2 - R_1} = n R_1$

Canceling $ R_1 $ from both sides:

$\frac{R_2}{R_2 - R_1} = n$

Rearranging for $ R_2/R_1 $, we get:

$\frac{R_2}{R_1} = n + n \frac{R_1}{R_2} \Rightarrow 1 + \frac{R_1}{R_2} = \frac{n}{n-1}$

Thus, the ratio of their radii is:

$\frac{R_2}{R_1} = \frac{n}{n-1}$

Therefore, the correct answer is:

$\frac{n}{n-1}$

Capacitance of an isolated conducting sphere of radius $R_1$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_2$ connected to earth The ratio of their radii $\left(\frac{ R _2}{ R _1}\right)$ is:

The Correct Option is A

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