Question

The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25 ✕ 105 V/m.What is the maximum energy that can be stored in the capacitor?

Answer 1

Explanation:

The given data is as follows.

        Dielectric constant, K = 3.0

   Area of the plates (A) = 0.021 $m^{2}$

   Distance between plates (d) = $2.75 \times 10^{-3} m$

  Maximum electric field (E) = $3.25 \times 10^{5} V/m$

Now, we will calculate the capacitance as follows.

             C = $\frac{k \epsilon_{o} \times A}{d}$

                 = $\frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}$

                 = $\frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}$

                 = $0.203 \times 10^{-9}$ F

Formula to calculate electric charge is as follows.

               E = $\frac{\sigma}{k \epsilon_{o}}$

or,           Q = $E \times k \times \epsilon_{o}A$      (as $\frac{\sigma}{\epsilon_{o}} = \frac{Q}{A}$)

                   = $3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021$

                   = $181.2 \times 10^{-9} C$

Formula to calculate the energy is as follows.

                  U = $\frac{1 \times Q^{2}}{2 \times C}$

                     = $\frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}$

                     = $\frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}$

                     = $9835.67 \times 10^{-9}$

or,                  = $98.35 \times 10^{-7} J$

Thus, we can conclude that the maximum energy that can be stored in the capacitor is $98.35 \times 10^{-7} J$.