Question

A variable capacitor is kept connected to a $10V$ battery. If the capacitance of the capacitor is changed from $7\mu F{\text{ }}to 3\mu F$, the change in energy of the capacitor is:

A. $2 \times {10^{ - 4}}{\text{ }}J$
B. $4 \times {10^{ - 4}}{\text{ }}J$
C. $6 \times {10^{ - 4}}{\text{ }}J$
D. $8 \times {10^{ - 4}}{\text{ }}J$

Answer 1

The energy stored on a capacitor or P.E. are often expressed in terms of the work done by a battery, where the voltage represents energy per unit charge. The voltage V is proportional to the quantity of charge which is already on the capacitor. Its expression is:
$Capacitor{\text{ }}energy{\text{ }}\left( E \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {capacitance} \right)\;{\left( {voltage} \right)^2}$

Complete step by step answer: Given that:

Battery{\text{ }}Voltage{\text{ }}\left( V \right) = {\text{ }}10V \\\ \begin{array}{*{20}{l}} {Capacitance{\text{ }}of{\text{ }}capacitor{\text{ }}initially{\text{ }}\left( {C_1} \right) = {\text{ }}7\mu F} \\\ {\;Capacitance{\text{ }}of{\text{ }}capacitor{\text{ }}after{\text{ }}\left( {C_2} \right){\text{ }} = {\text{ }}3\mu F} \end{array} \\\

A capacitor is a device to store energy. The process of charging up a capacitor involves the transferring of electric charges from one plate to a different.
The work wiped out in charging the capacitor is stored as its electrical P. E.
We will first find energy at initial when capacitance of capacitor $(C_1)$ was $7\mu F{\text{ }}$: $Capacitor{\text{ }}energy{\text{ }}\left( E \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {capacitance} \right)\;{\left( {voltage} \right)^2}$
Where C = Capacitance
V= Voltage

{Capacitor{\text{ }}energy{\text{ }}at{\text{ }}initially{\text{ }}\left( {E_1} \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {C_1} \right)\;{{\left( V \right)}^2}} \\\ {Capacitor{\text{ }}energy{\text{ }}at{\text{ }}initially{\text{ }}\left( {E_1} \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {7 \times {{10}^{ - 6}}} \right){{\left( {10} \right)}^2}{\text{ }} = {\text{ }}3.5\; \times {{10}^{ - 4}}{\text{ }}J} \end{array}$$ Now let’s find energy after change when capacitance of capacitor ($C_2$) becomes 3μ F: $$Capacitor{\text{ }}energy{\text{ }}\left( E \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {capacitance} \right)\;{\left( {voltage} \right)^2}$$ Where C = Capacitance V= Voltage $$\begin{array}{*{20}{l}} {Capacitor{\text{ }}energy{\text{ }}at{\text{ }}initially{\text{ }}\left( {E_2} \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {C_2} \right)\;{{\left( V \right)}^2}} \\\ {Capacitor{\text{ }}energy{\text{ }}at{\text{ }}initially{\text{ }}\left( {E_2} \right){\text{ }} = {\text{ }}1/2{\text{ }}\left( {3 \times {{10}^{ - 6}}} \right){{\left( {10} \right)}^2}{\text{ }} = {\text{ 1}}.5\; \times {{10}^{ - 4}}{\text{ }}J} \end{array}$$ change in energy of the capacitor when capacitance of the capacitor is changed from $$7\mu F {\text{ }}to3\mu F$$

Change{\text{ }}in{\text{ }}Energy{\text{ }}\left( {\Delta E} \right){\text{ }} = {\text{ }}E_1 - {\text{ }}E_2{\text{ }} = {\text{ }}3.5 \times ;{10^{ - 4}}{\text{ }}J{\text{ }} - {\text{ }}1.5 \times ;{10^{ - 4}}J{\text{ }} = {\text{ }}2 \times {10^{ - 4}}{\text{ }}J \\
Change{\text{ }}in{\text{ }}Energy{\text{ }}\left( {\Delta E} \right){\text{ }}is{\text{ }}2; \times {10^{ - 4}}J \\

$$∴ Option (A) is Correct. **Note:** The voltage (V) in the capacitor directly is proportional to the amount of charge which is already on the capacitor. The energy stored in the capacitor is directly proportional to the capacitor and directly proportional to the square of the voltage. Since the voltage is unchanged, the change in energy stored is directly proportional to the capacitor only. So, the change in energy is solely caused by change in the underlying capacitor.$$