Question

The capacitance of a fully-charged capacitor is 11 F. Determine the capacitor's capacitance when it is half charged.

Answer 1

Hi there!

`\boxed{C = 11F}`

Even though C = Q/V, the capacitance is NEVER changed by the charge or voltage.

The only factors that change the capacitance of a capacitor are those related to its geometry or if a dielectric is inserted. We can look at some examples:

For a parallel plate capacitor:

`C = (\epsilon_0A)/(d)`

C = Capacitance (F)

A = Area of plates (m²)
d = distance between plates (m)

For a spherical capacitor:

`C = 4\pi \epsilon_0 ((r_(outer)r_(inner))/(r_(outer) - r_(inner))})`

Notice how the capacitance is strictly determined by its geometric properties. Therefore, changing the charge or voltage has no effect, so the capacitance will remain 11 F.

Answer 2

The capacitance is 11 F for half and fully charged capacitor.

Step-by-step explanation:

Capacitance of capacitor is given by the expression

`C=(\epsilon A)/(d)`

Where ε is the vacuum permittivity, A is the area of plates and d is the separation between plates.

So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.

Here the capacitance is 11 F for half and fully charged capacitor.