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Q.
Four identical capacitors are connected in series with a $10\, V$ battery as shown in the figure. Potentials at $A$ and $B$ are
Electrostatic Potential and Capacitance
Solution:
If $q$ be charge on each capacitor, then
$\frac{q}{C}+\frac{q}{C}+\frac{q}{C}+\frac{q}{C}=10$ or
$\frac{q}{C}=\frac{10}{4}=2.5 V$
Now, $V_{A}-V_{N}=\frac{q}{C}+\frac{q}{C}+\frac{q}{C}$
$=\frac{3 q}{C}=3 \times 2.5$
$V_{A}-0=7.5, V_{A}=7.5 V$
Again, $V_{N}-V_{B}=\frac{q}{C}$
$0-V_{B}=2.5$ or $V_{B}=-2.5\, V$
