# Spherical 'capacitor' with only one charged plate

A question I came across asks about two concentric spherical conductors where only the inner sheet is charged. The question first asks to find the capacitance of such an arrangement. The first part of this question seems relatively easy. I can find the capacitance of such a capacitor by first finding the voltage difference between the two plates. Since the capacitance of a capacitor is determined only by the geometric configuration of the capacitor, this should be possible. However, after finding the voltage, to find the capacitance I would usually use the formula:

C = Q/V

Does this still work when q refers to the charge on just one plate? Also does this sort of a capacitor actually store energy. If so can the equation

1/2 CV^2

be used to calculate it. The final part of the question asks you to find the time taken for the voltage across the capacitor to fall to half it's original value when connected in series to a circuit with a resistor. Can the RC equations like:

q = Q(e^-t/RC)

still be used. Also, the question includes the gives the charge on the capacitor. Is this not extraneous information, since neither R nor C are affected by it? Extra points for anyone who shed some light on the physical/intuitive meaning of what is going on here?

If the inner conducting shell is charged then the charges will reside on the outside of the inner shell as there can be no electric field inside a conducting shell.

The charges on the outside of the inner conducting shell will produce a radial electric field and the outer conducting shell will find itself in that electric field.

However the outer conducting shell cannot have an electric field inside it so charges are induced on the surface of the outer conducting shell equal in magnitude to the charge on inner conducting shell.

If the charge on the inner shell is $+q$ then the charge on the inside of the outer shell will be $-q$ and the charge on the outside will be $+q$ if is is not earthed or zero if it is earthed.

So you have the normal arrangement for a capacitor with the charge $q$ in the equation $q=CV$ referring to the magnitude of the charge on one of the plates and your solution should reflect that.

You might not need the magnitude of $q$ for the last part of the problem but you did need it for the other parts?

• Thanks very much for the answer Farcher. Just to clarify, the charge should not need to be known to calculate how long it takes the capacitor to discharge half its voltage? – Padraig Stapleton Apr 29 '16 at 10:49
• Question link is missing? – Farcher Apr 29 '16 at 10:50
• Two concentric cylindrical conducting shells of length L are separated by a vacuum. The inner shell has surface charge density +σ and radius ra. The outer shell has radius rb. Using Gauss’ Law, as a function of radius r: Find an expression for the voltage and the electric field between the shells. Find the capacitance (d) The capacitor of part (b) is wired in series to a resistor R and allowed to discharge. If rb = 2 ra = 20 mm, L = 50 mm , σ = 20 nC m-2, and R = 5000 Ω, how long does it take for the voltage on the capacitor to reach half the original value? – Padraig Stapleton Apr 29 '16 at 10:53
• Just for clarity, "there can be no electric field inside a conducting shell. " means no electric field in the $(R, R+dr)$ interval (not in the $r<R$ one), right? – LLlAMnYP Apr 29 '16 at 11:44
• @LLlAMnYP There is no electric field inside a conductor however there is an electric field in the vacuum between the two conducting shells. – Farcher Apr 29 '16 at 12:38