Question

A parallel plate capacitor has a plate area of 1m2. The plates are separated by 2 mm, and the space between them is filled with jelly with a dielectric constant of 5.6. How much electrical energy is stored when the potential difference between the plates is 6000 V?Group of answer choices0.45 J90 J9,000 J45,000 J

Answer 1

Given,

The area of the parallel plate capacitor is

$A=1m^2$

The plates are separated by d = 2 mm.

The dielectric constant is

$\epsilon_r=5.6$

The capacitance of the capacitor is calculated by the formula.

$C=\frac{\epsilon_0\epsilon_rA}{d}$

Substitute the given values,

$\begin{gathered} C=\frac{8.85\times10^{-12}\times5.6\times1m^2}{2mm} \\ C=\frac{49.56\times10^{-12}}{2\times10^{-3}\text{ }} \\ C=24.78\times10^{-9}\text{ F} \end{gathered}$

The electrical energy stored in the plates is calculated by the formula.

$U=\frac{1}{2}CV^2$

The potential difference between the plates is 6000 V.

$\begin{gathered} U=\frac{1}{2}\times24.78\times10^{-9}\times(6000)^2 \\ U=2.59\times10^{-9}\times36\times10^6 \\ U=446.04\times10^{-3}\text{ }J \\ U=0.45\text{ J} \end{gathered}$

Thus, the correct option is 0.45 J.