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(B) Let $q$ be the charge on each capacitor. Since they are in series,the charge $q$ is the same for all.The total potential difference across the fou

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$7.5$

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(B) Let $q$ be the charge on each capacitor. Since they are in series,the charge $q$ is the same for all.The total potential difference across the four capacitors is $10 \, V$. Therefore,$\frac{q}{C} + \frac{q}{C} + \frac{q}{C} + \frac{q}{C} = 10 \, V$,which gives $\frac{q}{C} = 2.5 \, V$.Point $X$ is earthed,so its potential $V_X = 0 \, V$.Looking at the circuit diagram,point $A$ is connected to the positive terminal of the battery,and there are three capacitors between point $A$ and point $X$. The potential difference between $A$ and $X$ is $V_A - V_X = \frac{q}{C} + \frac{q}{C} + \frac{q}{C} = 3 	imes 2.5 \, V = 7.5 \, V$.Since $V_X = 0 \, V$,we have $V_A = 7.5 \, V$.

Four identical capacitors are connected in series with a battery of $emf$ $10 \, V$. The point $X$ is earthed. Find the potential of point $A$. (in $, V$)

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