Question

Two conducting spheres of radii R_{1} and R_{2} are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series−parallel connections.

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Solution

We need to calculate the capacitance of an isolated charged sphere.

Let us assume that the charge on the sphere is Q and its radius is R.

Capacitance of the charged sphere can be found by imagining a concentric sphere of infinite radius consisting of −Q charge.

Potential difference between the spheres = $\frac{1}{4\pi {\in}_{0}}\frac{Q}{R}$ − 0 = $\frac{1}{4\pi {\in}_{0}}\frac{Q}{R}$

Capacitance is the ratio of the magnitude of the charge on each conductor to the potential difference between them.

$C=\frac{Q}{{\displaystyle \frac{1}{4\pi {\in}_{0}}\frac{Q}{R}}}=4\pi {\in}_{0}R$

Therefore, the capacitances of spheres of radii R_{1} and R_{2} are C_{1} and C_{2}, respectively. They are given by

${C}_{1}=4\pi {\in}_{0}{R}_{1}\phantom{\rule{0ex}{0ex}}{C}_{2}=4\mathrm{\pi}{\in}_{0}{R}_{2}$

If the spheres are connected by a metal wire, the charge will flow from one sphere to another till their potentials become the same.

As there potentials become the same, the potential difference between the conductors for both the capacitors also becomes the same. Thus, it can be concluded that the capacitors are connected in parallel.

Thus, the capacitance of the combination is given by

${C}_{\mathrm{eq}}$ = C_{1} + C_{2}

$=4\mathrm{\pi}{\in}_{0}\left({R}_{1}+{R}_{2}\right)$

Let us assume that the charge on the sphere is Q and its radius is R.

Capacitance of the charged sphere can be found by imagining a concentric sphere of infinite radius consisting of −Q charge.

Potential difference between the spheres = $\frac{1}{4\pi {\in}_{0}}\frac{Q}{R}$ − 0 = $\frac{1}{4\pi {\in}_{0}}\frac{Q}{R}$

Capacitance is the ratio of the magnitude of the charge on each conductor to the potential difference between them.

$C=\frac{Q}{{\displaystyle \frac{1}{4\pi {\in}_{0}}\frac{Q}{R}}}=4\pi {\in}_{0}R$

Therefore, the capacitances of spheres of radii R

${C}_{1}=4\pi {\in}_{0}{R}_{1}\phantom{\rule{0ex}{0ex}}{C}_{2}=4\mathrm{\pi}{\in}_{0}{R}_{2}$

If the spheres are connected by a metal wire, the charge will flow from one sphere to another till their potentials become the same.

As there potentials become the same, the potential difference between the conductors for both the capacitors also becomes the same. Thus, it can be concluded that the capacitors are connected in parallel.

Thus, the capacitance of the combination is given by

${C}_{\mathrm{eq}}$ = C

$=4\mathrm{\pi}{\in}_{0}\left({R}_{1}+{R}_{2}\right)$

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