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=(\epsilon_0\epsilon_rA)/(d)`

Substitute the given values,

`\begin{gathered} C=(8.85*10^(-12)*5.6*1m^2)/(2mm) \\ C=\frac{49.56*10^(-12)}{2*10^(-3)\text{ }} \\ C=24.78*10^(-9)\text{ F} \end{gathered}`

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

`U=(1)/(2)CV^2`

The potential difference between the plates is 6000 V.

`\begin{gathered} U=(1)/(2)*24.78*10^(-9)*(6000)^2 \\ U=2.59*10^(-9)*36*10^6 \\ U=446.04*10^(-3)\text{ }J \\ U=0.45\text{ J} \end{gathered}`

Thus, the correct option is 0.45 J.