# Relation between charge and voltage in spherical capacitors

1. Feb 13, 2014

### abdo799

1-If we had a spherical capacitor and the voltage across it is 1000 V, I need to know the charge on every plate, I know 2 ways to solve this either using C=Q/V or V=Q/r * 1/4ε0∏. So, which one should I use, and why?
2-Another thing, according to hyperphysics the formula for the capacitance of the spherical capacitor = 4∏ε0/ (1/a - 1/b). Does that mean that the dielectric constant of the dielectric has no effect like parallel plates?

2. Feb 14, 2014

### BvU

2. Usually we look at spheres with nothing in between, hence the $\epsilon_0$. With a dielectric you would have to use $\epsilon$.

1. With plate you mean spherical surface ? In your case I would use the first eqn. Since you already found the hyperphysics expression for C: just above there is V in terms of Q, though. A bit more straightforward.

All this assuming you are comfortable with Gauss' law.

3. Feb 14, 2014

### abdo799

ok, i know from my other post ( chemical engineering thing) that you are a physicist which is really cool because i wanted to be one, about your answer, is the second wrong if i used it in this case?
if you considered a van de graaff generator , where the sphere was one plate and then it was covered with dielectric then we put another metal sphere on it as the other plate ( we ground this plate).(forming a spherical capacitor)
If the van de graaff generator had a radius of 1 m and put a charge on the sphere of 0.001 coulomb using the second equation the voltage on the sphere will be about 9 MV.
will the other plate gain the same voltage or the same charge ?
In my first case i was almost sure that we have to use C=Q/V because the setup was a capacitor and a battery, a classical circuit. In the case i just wrote , something is different, we charge only one plate with positive charge ( the van-de graaff generator) and the other plate will gain the extra negative charge from the ground.

Last edited: Feb 15, 2014
4. Feb 14, 2014

### tiny-tim

hi abdo799!
for a spherical capacitor with dielectric constant ε:

D = -Q/4πr2

E = -Q/ε4πr2

V = ∫ab E dx = [Q/ε4πr]ab

C = Q/V​

5. Feb 15, 2014

### abdo799

6. Feb 15, 2014

### tiny-tim

hi abdo799!
if the capacitor is part of a circuit, it's the same (and opposite) charge

when you change the voltage across any capacitor in a circuit, the total charge on the two plates will be the same as before

if the capacitor is not part of a circuit (ie it's not really acting as a capacitor), eg if you charge one sphere and bring it near an uncharged sphere (unconnected to it), then the positive and negative charges on the uncharged sphere will redistribute so that the opposite charge faces the first sphere, and the "same" charge retreats to the other side

of course, if the first plate surrounds the second so that the distance between them is constant, then there's no "nearer" and "further", and the charge in the second plate will have no reason to redistribute itself (it can't retreat to the inside surface of the second plate because the charge there must be zero from gauss's law since there's no charge inside it)

if the second plate surrounds the first, a gaussian surface in the middle of the second plate shows that the opposite charge on the inside surface must be equal to that of the inner plate: the "same" charge will retreat to the outside surface of the second plate if not grounded, and to the ground if grounded (i think that's what you're asking about?)

7. Feb 15, 2014

### abdo799

After we charge the whole thing , and the external sphere gained the extra charge , then we connect both sphere with a wire ( kinda like a capacitor discharge) the electrons will flow from the negative sphere to the positive one ,can i consider it now a capacitor ? can i use (1/2 c V^2) to calculate energy ?

8. Feb 16, 2014

### sophiecentaur

What value of C are you suggesting should be used?

9. Feb 17, 2014

### abdo799

C=4ε∏/(1/a-1/b)