Consider a capacitor circuit charged by a battery to a potential difference \( V_0 \) as shown in the figure.
When a dielectric slab is inserted into both the capacitors, new potential difference will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{8V_0}{5K} \)
The capacitors \( C_0 \) and \( 3C_0 \) are in parallel, and they are charged and then left isolated from the battery, so the total charge stays constant when the slabs are inserted.
Charge on the combination before the slabs are inserted:
\( \displaystyle q = (C_0 + 3C_0)\,V_0 = 4C_0 V_0 \)
A dielectric of constant \( K \) multiplies a capacitance by \( K \). From the figure, \( C_0 \) gets a slab of constant \( K \) and \( 3C_0 \) gets a slab of constant \( K/2 \):
\( \displaystyle C' = K C_0 + \frac{K}{2}\times3C_0 = \frac{5K C_0}{2} \)
The new potential difference is
\( \displaystyle V = \frac{q}{C'} = \frac{4C_0 V_0}{5KC_0/2} = \frac{8V_0}{5K} \)