# The total charge finally accumulated on all the spheres

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The given circuit consists of an ideal battery , n identical resistors each of resistance R and n+1 identical conducting spheres each of radius r . Assume the balls to be at great distance from each other as well from from the circuit . Find the total charge finally accumulated on all the spheres after the key is closed .

Unable to find the potential of the other side of the capacitance

asked Jul 9, 2018 1 flag
Each sphere has the same capacitance $C=4\pi\epsilon_0 R$. The potential difference across the capacitor determines the charge it stores. One plate is the surface of the sphere. The potential here is determined by the resistor circuit. Where is the other plate? What is the potential at the other plate? See my answer to your last question.

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All capacitors have two plates. For the spherical capacitor one plate is the surface of the sphere. In theory the 2nd plate is a concentric conducting shell of infinite radius. In practice the 2nd plate is all the distant surrounding objects, which are connected to the Earth. This looks very different from the concentric conducting shell at infinity, but practically it has approximately the same effect, provided that the closest objects are very much further than the diameter of the spherical capacitor.

So the 2nd plate can be considered to be grounded.

The potential difference across the battery is divided equally across the $n$ resistors, so the potential of each conducting sphere is an arithmetic progression $V_k=k\frac{V_0}{n}$ where $k=0, 1, 2...n$. Since the 2nd plate of each capacitor is grounded this is also the PD across each capacitor.

The charge stored on each capacitor is $Q_k=CV_k$ where $C=4\pi\epsilon_0 R$. The total charge stored on all capacitors is $$Q=(0+1+2+3+...+n)\frac{V_0}{n}C=\frac12 (n+1)CV_0$$ in which I have used the fact that the sum of the 1st $n$ numbers is $\frac12 n(n+1)$.

answered Jul 12, 2018 by (26,660 points)
edited Jul 13, 2018