Question

Suppose two parallel-plate capacitors have the same charge Q. but the area of capacitor 1 is A and the area of capacitor 2 is 2A.
1. If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2?

Answer 1

Answer:

V/2

Explanation:

By definition, the capacitance is the quotient between the charge on one of its plates and the voltage between them:

C = Q/V

For a parallel-plate capacitor, applying Gauss' Law to a closed surface enclosing one of the plates, it can be showed that the value of the capacitance depends only of the geometry, and the material that fills the space between plates, as follows:

C = ε*A/d, where ε= dielectric constant of the material between plates, A is the area of the capacitor and d the distance between plates.

Let's call  C₁, to the capacitor of area A, and C₂, to the one of area 2*A.

So, we can write the following expression:

C₂ = ε*2*A / d = 2* (ε*A/d) = 2* C₁

If Q remains constant, we can write the following equality:

⇒ Q = C₁*V = 2*C₁*V₂

Solving for Vx, we have:

V₂ = V/2

if the value of the capacitance is doubled, in order to keep Q constant, V₂ must be half of V.

Answer 2

Hi there!

Recall the following:

$C = \frac{Q}{V}$

C = Capacitance (F)
Q = Charge (C)

V = Voltage (V)

Remember that changing the charge or voltage does NOT change the capacitance, but rather properties (Area of plate or plate distance) of the capacitor must change for the relationship to stay true.

The capacitance of a parallel plate capacitor can be found using the following:

$C = \frac{\epsilon_0A}{d}$

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

Since capacitor 2 has an area of 2A, using the above relation:
$C' = \frac{\epsilon_0(2A)}{d} = 2C$

The capacitance is doubled, so something about Q or V must change.

Since Q is held constant (stated in the question), the VOLTAGE must change.

To mirror the doubled capacitance, Q/V must be doubled. We find that we can HALF the voltage to do this.

$\frac{Q}{\frac{1}{2}V}} = 2\frac{Q}{V}$

Thus, the voltage across capacitor 2 is V/2.