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In a spherical capacitor, we have two concentric spherical shells, the inner one carrying a charge $Q$ and the outer one carrying charge $-Q$. If the inner shell is displaced from the center without touching the other shell then will the capacitance increase or decrease?

I was trying to solve this question and my solution is that we do work on the capacitor to move the inner shell so the potential energy of the capacitor should increase. Using $U = Q^2/2C$, capacitance should decrease. But the answer is that capacitance increases. Where is the flaw in my logic? Is there any other solution to this problem?

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  • $\begingroup$ How do you know you are doing work (with a positive sign) on the capacitor? What about your formula U = Q^2/2C? Typically, U stands for a voltage, and then it should be U = Q/C. So, what does your U stand for? $\endgroup$ Jun 24, 2021 at 7:25
  • $\begingroup$ U is potential energy or energy $\endgroup$
    – Niescte
    Jun 24, 2021 at 8:19

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Using Gauss' law we know that for a spherically symmetric charge distribution the electric outside is that of a point charge at its centre with the same charge. Consider the set-up:

enter image description here

On the left is the undisplaced situation for which the spherically symmetric case holds for the whole setup. On the right is the system once the inner sphere has been displaced by $d$.

Now consider fixing the total charge on each sphere in the initial state (for example disconnecting the capacitor plates). But addition fixing the charge distribution of the inner sphere such that it remains spherically symmetrically about the centre of the inner sphere (treating the inner sphere temporarily like an insulator). I will later justify this step as it will become clearer.

Next, displace the inner sphere by $d$. The charge on the outer sphere will redistribute to maintain an equipotential surface - but as stated above the inner sphere's charge distribution is held constant. Because the inner sphere's charge distribution is held constant we can use the method of images to replace the inner sphere with a point charge at its centre with the same charge $Q$. Now we can apply the method of images to replace the outer sphere with another point charge $-Q\frac{R}{d}$:

enter image description here

where the blue points are point charges. Thus, the voltage between the inner sphere and the outer sphere is:

$$V=\frac{Q}{4\pi\epsilon\epsilon_0}\left(\frac{1}{r}-\frac{R}{R^2-rd-d^2}\right)$$

enter image description here

As we can see the voltage reduces as $d$ increases; however, we still haven't accounted for the redistribution of charge on the inner sphere. So now we release the fixed charge on the inner sphere and allow it to redistribute. Now while it is difficult to now calculate exactly the voltage in this case we know it must be less than the voltage calculated above because the charge redistributes until the conductor is an equipotential surface and so minimises the potential energy. So now both the initial and final states are physical so it does not matter that we took the non-physical root of fixing the charge distribution along the way, this was just a logical convenience.

Hence, the voltage $V$ has decreased while the charge $Q$ has remained constant. As the capacitance $C\equiv\frac{Q}{V}$ then the capacitance increase.

Potential Energy Approach

As the potential decreases, the potential energy decreases as $d$ increases and so the undisplaced case is an unstable equilibrium. The decrease in potential energy means $C$ must increase as $U=\frac{Q^2}{2C}$.

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    $\begingroup$ Nice answer, but since the charged sphere is displaced , wont the the charge distribution cease to be spherically symmetric on the sphere. It will be more concentrated at points close to the bigger sphere and less away from it. Then how can you use the formula V = KQ/r. $\endgroup$
    – Niescte
    Jun 24, 2021 at 11:31
  • $\begingroup$ You mentioned in your answer that we fix the charge . How do we do that ? Isn't it contrary to what would happen in the actual scenario ? $\endgroup$
    – Niescte
    Jun 24, 2021 at 11:33
  • $\begingroup$ My bad, I believe you are correct that it would redistribute. $\endgroup$
    – Chris Long
    Jun 24, 2021 at 11:43
  • $\begingroup$ However, we can fix the charge as the capacitance is a property of the two plates regardless of whether they are connected or not. Thus, for mathematical convince, we can leave them disconnected. $\endgroup$
    – Chris Long
    Jun 24, 2021 at 11:45
  • $\begingroup$ You can answer this using method of images instead and that should account for redistribution, I will rework my answer. $\endgroup$
    – Chris Long
    Jun 24, 2021 at 11:54

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