I just walked out of a physics test, and one of the questions was to the effect of:

You charge a parallel plate capacitor to 'Q', disconnect the battery, and double the distance between the plates. How does the potential difference (Voltage) change?

After some thought, I deduced that if Q remains the same and Q=CV (C=capacitance in farads), then V must double to maintain the same charge Q.

My reasoning is that by separating the plates, you are doing work on them. This energy must be stored somewhere, and since C cannot increase (In fact it is dropping) then V must increase by doubling.

The overall energy in the capacitor will increase according to U=C(V^2)/2, so the energy will double after accounting for the drop in capacitance.

Sound correct?
If I miss this question I don't have the luxury of an explanation as to why I missed it without going to some inconvenient office hours.

 

You have done well here and deserve a pint.

If you wish to do it mathematically , you have not used the fact that the capacitance between two plates is proportional to the area and inversely proportional to the distance apart, the constant being the permittivity.


\(C = \frac{{\varepsilon A}}{d}\)

Using this to obtain the ratio of capacitance as distance, d doubles


\(\frac{{{C_2}}}{{{C_1}}} = \frac{{\frac{{\varepsilon A}}{{2{d_1}}}}}{{\frac{{\varepsilon A}}{{{d_1}}}}} = \frac{1}{2}\)

or


\({C_1} = 2{C_2}\)

Using energy as an intermediary as you have done, at constant charge, Q

\(U = \frac{1}{2}\frac{{{Q^2}}}{C}\)

So the ratio of energies is using

\(U = \frac{1}{2}\frac{{{Q^2}}}{C}\)

\(\frac{{{U_2}}}{{{U_1}}} = \frac{{\frac{{{Q^2}}}{{2{C_2}}}}}{{\frac{{{Q^2}}}{{4{C_2}}}}} = 2\)

or

\({U_2} = 2{U_1}\)

Inserting the ratio of energies and using your different energy relation


\(\frac{{{U_2}}}{{{U_1}}} = \frac{{\frac{{Q{V_2}}}{2}}}{{\frac{{Q{V_1}}}{2}}} = 2\)

or


\({V_2} = 2{V_1}\)

This confirms all the ratios you deduced.

 

I just walked out of a physics test, and one of the questions was to the effect of:

You charge a parallel plate capacitor to 'Q', disconnect the battery, and double the distance between the plates. How does the potential difference (Voltage) change?

After some thought, I deduced that if Q remains the same and Q=CV (C=capacitance in farads), then V must double to maintain the same charge Q.

My reasoning is that by separating the plates, you are doing work on them. This energy must be stored somewhere, and since C cannot increase (In fact it is dropping) then V must increase by doubling.

The overall energy in the capacitor will increase according to U=C(V^2)/2, so the energy will double after accounting for the drop in capacitance.

Sound correct?
If I miss this question I don't have the luxury of an explanation as to why I missed it without going to some inconvenient office hours.

Very good job!

Your explanation is well-reasoned and easy to follow.

About all that could be done more would be to show it mathematically (which studiot has already done).

 

I believe this is included in the Kasmir (spelling) effect.
cliff

 

I believe this is included in the Kasmir (spelling) effect.
cliff

No this is just bog standard electric circuit theory in classical physics.

The casimir effect is a quantum effect.

http://en.wikipedia.org/wiki/Casimir_effect

 

An real life application of this result is the condenser microphone.