Capacitance, voltage, distance and charge?

[seto6]

NOT HOME WORK QUESTION!

If a capacitor is attached to a battery, initially the separation between the capacitor plates is D then the distance is doubled(2D). The voltage should remain the same because it is still attached to a battery. C=E0A/D so the capacitance decrease by factor of 2. my question is that since the capacitance decreased, the voltage or the charge must have decreased. so did the charge decrease or increase?

 

[graphene]

V remains the same, the charge on the capacitor halves.

 

[uart]

my question is that since the capacitance decreased, the voltage or the charge must have decreased. so did the charge decrease or increase?

Yes the charge on the capacitor certainly did decrease. Exactly half of the initial charge did in fact flow back into the battery.

It's interesting to calculate the electrical energies involved in this case. It's relatively easy to show that a stored energy of $\frac{1}{4} C_0 V^2$ Joules was lost by the capacitor while a net energy of $\frac{1}{2} C_0 V^2$ Joules was returned to the battery.

I'll leave it as an exercise to figure out where the other $\frac{1}{4} C_0 V^2$ J of energy came from.

 

[clombard1973]

NOT HOME WORK QUESTION!

If a capacitor is attached to a battery, initially the separation between the capacitor plates is D then the distance is doubled(2D). The voltage should remain the same because it is still attached to a battery. C=E0A/D so the capacitance decrease by factor of 2. my question is that since the capacitance decreased, the voltage or the charge must have decreased. so did the charge decrease or increase?

Charge, voltage and capacitance are related by the equation Q = VC, where Q is the charge on the plates of the capacitor, C is the capacitance of the capacitor, and V is the voltage dropped across the capacitor.

So if you double the plate separation distance, and you know that the capacitance is inversely related to that distance, then you must have cut the capacitance by half.

C = E_OE_R(A/d) and if d is now 2d then C' = E_OE_R(A/2d) = (1/2)C, so the new capacitance C' is 1/2 the old capacitance, C.

So, now plug that back into the function relating the voltage and capacitance to the charge on the plates. Let Q be the charge that was on the plates when the capacitor plates were separated by a distance, d, and let Q' be the charge on the plates by separating them by twice the distance:

Q = VC
Q' = V(1/2)C = (1/2)VC
Q' = (1/2)Q

The amount of charge on the plates by separating them by twice the distance they were separated at causes the charge on the plates to drop by half also.

Many Smiles,
Craig

 

[sardar]

I would say that the charge must have decreased in this scenario. This is because the capacitance is directly proportional to the charge, meaning that as the capacitance decreases, the charge must also decrease. This is due to the fact that the capacitance is a measure of the ability of a capacitor to store charge, so as the capacitance decreases, the ability to store charge also decreases. Therefore, the charge must have decreased in order to maintain a constant voltage.