Consider the figure shown. The potential difference $V_{A} - V_{B}$ is :

E [\frac{C_{1} C_{4} - C_{2} C_{3}}{(C_{1} + C_{2}) (C_{3} + C_{4})}]

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E [\frac{C_{1} C_{4} + C_{2} C_{3}}{(C_{1} + C_{3}) (C_{2} + C_{4})}]

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E [\frac{C_{1} C_{3} - C_{2} C_{4}}{(C_{1} + C_{2}) (C_{3} + C_{4})}]

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E [\frac{C_{1} C_{3} - C_{2} C_{4}}{(C_{1} + C_{3}) (C_{2} + C_{4})}]

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Solution

The correct option is A $E [\frac{C_{1} C_{4} - C_{2} C_{3}}{(C_{1} + C_{2}) (C_{3} + C_{4})}]$
Let the charge be as shown. (Capacitors in series have the same charge)
Taking loop containing $C_{1}, C_{2}$ and $E$ we get
$\frac{q}{C_{1}} + \frac{q}{C_{2}} - E = 0 o r q = E [\frac{C_{1} C_{2}}{C_{1} + C_{2}}]$
Similarly, from loop containing $C_{3}, C_{4}, E$ we get
$\frac{q^{'}}{C_{3}} + \frac{q^{'}}{C_{4}} - E = 0 o r q^{'} = E [\frac{C_{3} C_{4}}{C_{3} + C_{4}}]$
Now,
$V_{A} - V_{B} = \frac{q}{C_{2}} - \frac{q^{'}}{C_{4}}$
$= E [\frac{C_{1}}{C_{1} + C_{2}} - \frac{C_{3}}{C_{3} + C_{4}}]$
$= E [\frac{C_{1} C_{4} - C_{3} C_{2}}{(C_{1} + C_{2}) (C_{3} + C_{4})}]$

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