Question

A parallel plate capacitor is made with a separation betweentheplates of 6.3×10−7m and the gap is filled with a dielectric with 휅=9.6. The capacitor is then connected to a battery with an emf of 400 V, andthe capacitorstores 120 J of energy in the electric field.What is the area of the capacitor plates?A.9.4×10-3m2B.9.0×10-2m2C.5.6×100m2D.1.1×101m2E.1.1×102m2

Answer 1

D)
`A=1.1*10^1m^2`

Step-by-step explanation:

The capacitance of a parallel plate capacitor with a separation between the plates d, area of the plates A filled with a dielectric with relative permittivity k is given by the formula:

`C=(k \epsilon_0 A)/(d)`

where
`\epsilon_0=8.85*10^(-12)F/m` is the permittivity of free space (vacuum).

The energy stored on a capacitor can be calculated with:

`U=(CV^2)/(2)`

Combining these two equations we have:

`U=(k \epsilon_0 A V^2)/(2d)`

Which for the area and our values is:

`A=(2dU)/(k \epsilon_0 V^2)=(2(6.3*10^(-7)m)(120J))/((9.6)(8.85*10^(-12)F/m)(400V)^2)=11.1m^2`