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Question: The work done in placing a charge of \[8 \times {10^{ - 18}}\] coulomb on a condenser of capacity 10...

The work done in placing a charge of 8×10−188 \times {10^{ - 18}} coulomb on a condenser of capacity 100 micro-farad is :
A. 3.1×10−26J B. 4×10−10J C. 32×10−32J D. 16×10−232J \begin{gathered} {\text{A}}{\text{. 3}}{\text{.1}} \times {\text{1}}{{\text{0}}^{ - 26}}J \\\ {\text{B}}{\text{. 4}} \times {\text{1}}{{\text{0}}^{ - 10}}J \\\ {\text{C}}{\text{. 32}} \times {\text{1}}{{\text{0}}^{ - 32}}J \\\ {\text{D}}{\text{. 16}} \times {\text{1}}{{\text{0}}^{ - 232}}J \\\ \end{gathered}

Explanation

Solution

The work done in placing a charge on a condenser is equal to the square of the charge divided by capacitance of the condenser. We have the value of charge and the capacitance, putting them in the formula for the work done, we can obtain the required answer.
Formula used:
The work done in placing a charge q on a condenser of capacitance C is given by the following expression:
W=q22CW = \dfrac{{{q^2}}}{{2C}}

Complete answer:
We are given the value of charge as
q=8×10−18Cq = 8 \times {10^{ - 18}}C
The capacitance of the given condenser has the following value:
C=100μF=100×10−6F=10−4FC = 100\mu F = 100 \times {10^{ - 6}}F = {10^{ - 4}}F
Now we need to find out the work done in placing the above charge on the condenser. It is given by the following formula :
W=q22CW = \dfrac{{{q^2}}}{{2C}}
Inserting the values of the known parameters, we get
W=q22C=(8×10−18)22×10−4=32×10−32JW = \dfrac{{{q^2}}}{{2C}} = \dfrac{{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}}}{{2 \times {{10}^{ - 4}}}} = 32 \times {10^{ - 32}}J

So, the correct answer is “Option C”.

Additional Information:
The capacitance of a capacitor is given in terms of the charge stored in the capacitor and the potential difference of the plates of the capacitor by the following expression.
q=CVq = CV
Using this expression, we can convert the expression for work done in terms of the voltage and the capacitance. Then the energy stored in the capacitor is given as
E=12CV2E = \dfrac{1}{2}C{V^2}
A capacitor is a device which stores electrical energy while an inductor is a device which stores magnetic energy.

Note:
1. A capacitor is commercially known as a condenser which acts like a source of stored electrical energy.
2. The capacitance of a capacitor is the measure of the ability of that capacitor to store electrical energy and tells us how much charge can be stored between two plates of the capacitor having certain potential differences.