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Consider a system of two conducting spheres, as follows

enter image description here

The outer sphere contains a charge $+Q$ and the inner surface is neutral.

What would be the capacitance of the system?
We know $C=\frac{q}{V}$. But here the potential difference between the spheres is 0, so will it be correct to say, that the capacitance of the system is infinity. but if I consider this whole system as a single conductor and see its potential difference with respect to infinity, I get $C=4\pi \epsilon _0(3R)$. Now suppose I earth the inner sphere. then to make the potential of inner sphere=$0$, $\frac{Q}{3}$ charge will flow into the earth. So now a potential difference will occur between the spheres. Then the capacitance of the system will not be $\infty$. and there will be a potential difference with respect to infinity. so how exactly can we calculate the capacitance? Please help me solve this doubt. I am really confused here. I think I have not understood what the term "capacitance" means exactly.

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  • $\begingroup$ You will have two capacitors in parallel. $\endgroup$
    – Farcher
    Commented Aug 20, 2021 at 18:55
  • $\begingroup$ If the inner sphere is uncharged, there is no potential difference between them. $\endgroup$ Commented Aug 20, 2021 at 21:27

2 Answers 2

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In general, for concentric spheres $(a<b)$,

$$ \begin{pmatrix} V_a \\ V_b \end{pmatrix}= \frac{1}{4\pi \epsilon_0} \begin{pmatrix} \frac{1}{a} & \frac{1}{b} \\ \frac{1}{b} & \frac{1}{b} \end{pmatrix} \begin{pmatrix} Q_a \\ Q_b \end{pmatrix}$$

  • The matrix is known as elastance matrix which is the inverse of the capacitance matrix.

The mutual capacitance is given by

$$C=\frac{Q_a}{V_a-V_b}=\frac{4\pi \epsilon_0ab}{b-a}$$

  • Note that there's no dependence on charges on the outer sphere due to Gauss' Law.

  • In usual practice, the outer sphere is earthed so as $V_b=0$.

See also another post of mine here.

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If you do not connect the inner sphere to ground or some other potential , the inner sphere does not matter you can choose any r for it, you just have the capacitance of the outer sphere .

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  • $\begingroup$ so, if i connect the inner sphere to ground the capacitance will be q/v where q is the charge on the outer surface of inner sphere. and if it is not connected to ground capacitance will be seen for the outer sphere wrt infinity. am i correct? $\endgroup$
    – Nimit Jain
    Commented Aug 20, 2021 at 17:29
  • $\begingroup$ yes you are right, but v depend of R and r. $\endgroup$
    – trula
    Commented Aug 20, 2021 at 20:43

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