Question

Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The capacitor is charged and stores 8.11 × 10 − 9 J of energy. Find the electric field strength E inside the capacitor.

Answer 1

Explanation:

It is known that the relation between electric field and potential is as follows.

             E = $\frac{V}{d}$

And, formula to calculate the capacitance is as follows.

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

              = $\frac{8.854 \times 10^{-12} \times (0.479 m)^{2}}{0.479 \times 10^{-3}}$

              = $4.24 \times 10^{-9}$ F

Hence, energy stored in a capacitor is as follows.

         W = $\frac{1}{2}CV^{2}$

          V = $\sqrt{\frac{2W}{C}}$

        E = $\sqrt{\frac{2W}{d^{2}C}}$

            = $\frac{2 \times 8.11 \times 10^{-9} J}{(0.479 \times 10^{-3})^{2} \times 4.24 \times 10^{-9}}$

            = $16.687 \times 10^{3} N/C$

Thus, we can conclude that electric field strength E inside the capacitor is $16.687 \times 10^{3} N/C$.