Question

A certain parallel plate capacitor is filled with a dielectric for which k = 5.5. The area of each plate is $\mathbf{0}\mathbf{.}\mathbf{034}{\mathit{m}}^{\mathbf{2}}$, and the plates are separated by 2.0 mm.The capacitor will fail (short out and burn up) if the electric field between the plates exceeds$\mathbf{200}\frac{\mathbf{k}\mathbf{N}}{\mathbf{C}}$. What is the maximum energy that can be stored in the capacitor?

Short answer

The maximum energy that can be stored in the parallel plate capacitor is$6.6\times {10}^{-5}J$ .

Step-by-step solution

The given data

a) Dielectric constant filled within the capacitor, k = 5.5

b) Area of each plate,$A=0.034m^2$

c) Separation between the plates,$d=2.0\times {10}^{-3}m$

d) Maximum electric field,$E=200\times {10}^{3}N/C$

Understanding the concept of the stored energy and capacitance

A capacitor consists of two isolated plates with charges +q and -q. The capacitance Cof the capacitor is defined from,

q = CV

Here,is the potential difference between the two plates.

Energy stored in a parallel plate capacitor is given by the relation. Before using this, we have to find the charge on the capacitor which can be calculated from the electric field expression using the surface density concept.

Formulae:

The surface density of a material,$\sigma =\frac{q}{A}$ …(i)

Here,$\sigma$ is surface charge density, q is charge and A is area of cross section.

The electric field within a body due to flux,$E=\frac{\sigma }{k.{\epsilon }_0}$ …(ii)

Here, E is electric field, k is dielectric constant of the material,${\epsilon }_0$ is permittivity of the free space.

The capacitance formula due to a present dielectric material, $C=\frac{k.{\epsilon }_0.A}{d}$ …(iii)

Here, C is capacitance, A is area of cross section, d is separation between the two plates.

The energy stored within the capacitor plates, $U=\frac{q^2}{2,C}$ …(iv)

Here, U is energy stored in the capacitor, q is charge stored in the capacitor.

Calculation of the maximum stored energy

Substituting equation (i) in equation (ii), we can get the charge present within the capacitor by using the given data as follows:

$$\begin{gathered} E=\frac{q}{k{\epsilon }_{0}A} \\ q=E.k.{\epsilon }_0.A \\ =\left(200\times {10}^{3}N/C\right)\times \left(5.5\right)\times \left(8.85\times {10}^{-12}{C}^2/N.m^2\right)\times \left(0.034m^2\right) \\ =3.3\times {10}^{-7}C \end{gathered}$$

Energy stored in the capacitor can be calculated using the given data and equation (iii) in equation (iv) as follows:

$$\begin{gathered} U=\frac{q^{2}d}{2k{\epsilon }_{0}A} \\ =\frac{{\left(3.3\times {10}^{3}C\right)}^2\times \left(2.0\times {10}^{-3}m\right)}{2\times 5.5\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)\times 0.034m^2} \\ =6.6\times {10}^{-5J} \end{gathered}$$

Hence, the value of the maximum energy that can be stored is$6.6\times {10}^{-5J}$ .