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

Answer:

D) $A=1.1\times10^1m^2$

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:

$C=\frac{k \epsilon_0 A}{d}$

where $\epsilon_0=8.85\times10^{-12}F/m$ is the permittivity of free space (vacuum).

The energy stored on a capacitor can be calculated with:

$U=\frac{CV^2}{2}$

Combining these two equations we have:

$U=\frac{k \epsilon_0 A V^2}{2d}$

Which for the area and our values is:

$A=\frac{2dU}{k \epsilon_0 V^2}=\frac{2(6.3\times10^{-7}m)(120J)}{(9.6)(8.85\times10^{-12}F/m)(400V)^2}=11.1m^2$