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Question: The capacitance of a variable capacitor joined with a battery of 100 V is changed from \(2\,{\rm{\mu...

The capacitance of a variable capacitor joined with a battery of 100 V is changed from 2 μF2\,{\rm{\mu F}} to 10 μF10\,{\rm{\mu F}}. What is the change in the energy stored in it?
(a) 2×10−2  J2 \times {10^{ - 2}}\;{\rm{J}}
(b) 2.5×10−2  J2.5 \times {10^{ - 2}}\;{\rm{J}}
(c) 6.5×10−2  J6.5 \times {10^{ - 2}}\;{\rm{J}}
(d) 4×10−2  J4 \times {10^{ - 2}}\;{\rm{J}}

Explanation

Solution

The capacitor is used to store energy that can be supplied whenever required. The capacitor's energy directly depends on the capacitance of the capacitor and the supplied voltage. To determine the change in energy stored, first, we will determine the initial energy stored by the capacitor and final energy stored by the capacitor separately, and then we will subtract the magnitude of the initial stored energy from the final stored energy.

Complete step by step answer:
It is given to us in the question that the initial capacitance of the variable capacitor is 2  μF{\rm{2}}\;{\rm{\mu F}} and the final capacitance of the variable capacitor is 10  μF10\;{\rm{\mu F}}.
The voltage of the battery is 100 V, so that we will use this information for the determination of the initial and final energy stored by the capacitor.
We will write the expression for the energy stored by the capacitor in the initial condition.
E1=12C1V2{E_1} = \dfrac{1}{2}{C_1}{V^2}
Here, C1{C_1} is the initial capacitance of the variable capacitor and VV is the supplied voltage.
On Substituting the values in the above expression. Therefore, we get,
E1=12(2  μF×1×10−6  F1  μF)(100  V)2 E1=1×10−2  J\begin{array}{l} {E_1} = \dfrac{1}{2}\left( {{\rm{2}}\;{\rm{\mu F}} \times \dfrac{{1 \times {{10}^{ - 6}}\;{\rm{F}}}}{{1\;{\rm{\mu F}}}}} \right){\left( {100\;{\rm{V}}} \right)^2}\\\ {E_1} = 1 \times {10^{ - 2}}\;{\rm{J}} \end{array}
Also, write the expression for the final energy stored by the capacitor.
E2=12C2V2{E_2} = \dfrac{1}{2}{C_2}{V^2}
Here, C2{C_2} is the final capacitance of the variable capacitor.
On Substituting the values in the above expression. Therefore, we get,
E2=12(10  μF×1×10−6  F1  μF)(100  V)2 E2=5×10−2  J\begin{array}{l} {E_2} = \dfrac{1}{2}\left( {10\;{\rm{\mu F}} \times \dfrac{{1 \times {{10}^{ - 6}}\;{\rm{F}}}}{{1\;{\rm{\mu F}}}}} \right){\left( {100\;{\rm{V}}} \right)^2}\\\ {E_2} = 5 \times {10^{ - 2}}\;{\rm{J}} \end{array}
Now we will subtract the initial stored energy form the final stored energy so that we can obtain the change in energy stored, we have,
E=E2−E1E = {E_2} - {E_1}
On Substituting the values in the above expression. Therefore, we get,
E=(5  ×10−2J)−  (1×10−2J) E=4×10−2  J\begin{array}{l} E = \left( {5\; \times {{10}^{ - 2}}{\rm{J}}} \right) - \;\left( {1 \times {{10}^{ - 2}}{\rm{J}}} \right)\\\ E = 4 \times {10^{ - 2}}\;{\rm{J}} \end{array}
Therefore, the change in the energy stored is 4×10−2  J4 \times {10^{ - 2}}\;{\rm{J}} and option (d) is correct.

Note: Do not make the calculation and solution complex by directly using the equationE=12C2V2−12C1V2E = \dfrac{1}{2}{C_2}{V^2} - \dfrac{1}{2}{C_1}{V^2}. Always determine the energy stored separately for the initial and final condition, so that the calculation and solution become easy and simple for the understanding.