0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ You will have two capacitors in parallel. $\endgroup$
    – Farcher
    Aug 20, 2021 at 18:55
  • $\begingroup$ If the inner sphere is uncharged, there is no potential difference between them. $\endgroup$ Aug 20, 2021 at 21:27

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 .

$\endgroup$
2
  • $\begingroup$ 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
    Aug 20, 2021 at 17:29
  • $\begingroup$ yes you are right, but v depend of R and r. $\endgroup$
    – trula
    Aug 20, 2021 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.