Question

Change in energy of the capacitor in above case is:

• A. $162\times {10}^{-8}J$
• B. $-1215\times {10}^{-10}J$
• C. $5.06\times {10}^{-8}J$
• D. $-303.75\times {10}^{-9}J$

Answer 1

Answer: (A)

$162\times {10}^{-8}J$

The energy stored in the capacitor is given as

$E=\frac{1}{2}C{V}^2$

$C=\frac{{\epsilon }_{0}A}{d}$

By substituting the values in the above equation we get

$C=\frac{9\times {10}^{-12}\times {10}^3}{2.5\times {10}^2}$

$C=36\times {10}^{-11}$

$E=\frac{1}{2}\times 36\times {10}^{-11}\times 300\times 300$

$162\times 10^{-8}J$