Question
If the potential of a capacitor having Capacity $$6\,\mu F$$ is increased from $$10\,V$$ to $$20\,V,$$ then increase in its energy will be
A.
$$4 \times {10^{ - 4}}\,J$$
B.
$$4 \times {10^{ - 14}}\,J$$
C.
$$9 \times {10^{ - 4}}\,J$$
D.
$$12 \times {10^{ - 6}}\,J$$
Answer :
$$9 \times {10^{ - 4}}\,J$$
Solution :
Energy stored in a charged capacitor is in the form of electric field energy and it resides in the dielectric medium between the plates. This energy stored in the capacitor is given by $$U = \frac{1}{2}C{V^2}$$
If initial potential is $${V_1}$$ and final potential is $${V_2},$$ then increase in energy $$\left( {\Delta U} \right)$$
$$\eqalign{
& \Delta U = \frac{1}{2}C\left( {V_2^2 - V_1^2} \right) \cr
& = \frac{1}{2} \times \left( {6 \times {{10}^{ - 6}}} \right) \times \left[ {{{\left( {20} \right)}^2} - {{\left( {10} \right)}^2}} \right] \cr
& = \left( {3 \times {{10}^{ - 6}}} \right) \times 300 \cr
& = 9 \times {10^{ - 4}}\,J \cr} $$