Question

Two metallic spheres of radii a and b are separated by a distance d as shown in figure. The capacity of the system is:

• A. $\frac{4\pi {\epsilon }_0}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$
• B. $\frac{4\pi {\epsilon }_0}{\frac{1}{a}+\frac{1}{b}+\frac{2}{d}}$
• C. $\frac{4\pi {\epsilon }_0}{\frac{1}{a}-\frac{1}{b}+\frac{2}{d}}$
• D. $\frac{4\pi {\epsilon }_0}{\frac{1}{a}-\frac{1}{b}-\frac{2}{d}}$

Answer 1

The correct option is A $\frac{4\pi {\epsilon }_0}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$

Potential $V_1=\frac{Q}{4\pi {\epsilon }_0}(\frac{1}{a}-\frac{1}{d})$

$V_2=\frac{Q}{4\pi {\epsilon }_0}(\frac{1}{d}-\frac{1}{b})$

So, the potential difference is,

$V_1-V_2=\frac{Q}{4\pi {\epsilon }_0}(\frac{1}{a}+\frac{1}{b}-\frac{2}{d})$. . . . .(1)

As we know,

Capacitance is $C=\frac{Q}{V_1-V_2}$

substitute equation (1) in the above equation,

$C=\frac{4\pi {\epsilon }_0}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$

The correct option is A.