Question

Four identical capacitors are connected in series with a $10V$ battery as shown in the figure. Potentials at $A$ and $B$ are

• A. 10 V , 0 V
• B. 7.5 V ,-2.5 V
• C. 5 V ,-5 V
• D. 7.5 V , 2.5 V

Answer 1

Answer: (B)

If $q$ be charge on each capacitor, then
$\frac{q}{C}+\frac{q}{C}+\frac{q}{C}+\frac{q}{C}=10$ or
$\frac{q}{C}=\frac{10}{4}=2.5V$
Now, $V_A-V_N=\frac{q}{C}+\frac{q}{C}+\frac{q}{C}$
$=\frac{3q}{C}=3\times 2.5$
$V_A-0=7.5,V_A=7.5V$
Again, $V_N-V_B=\frac{q}{C}$
$0-V_B=2.5$ or $V_B=-2.5V$