Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively
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(b) ${C_{eq}} = \frac{{C \times 4C}}{{(C + 4C)}} = \frac{{4C}}{5}$
$Q = {C_{eq}}.V = \frac{{4C}}{5} \times 100 = 80\,C$
Hence ${V_1} = \frac{Q}{{{C_1}}} = \frac{{80C}}{{{C_1}}} = 80\,V$
and ${V_2} = \frac{{80C}}{{4C}} = 20\,V$
$Q = {C_{eq}}.V = \frac{{4C}}{5} \times 100 = 80\,C$
Hence ${V_1} = \frac{Q}{{{C_1}}} = \frac{{80C}}{{{C_1}}} = 80\,V$
and ${V_2} = \frac{{80C}}{{4C}} = 20\,V$
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