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CGP EDU Academic Team
Published on: September 13, 2026
Two identical parallel plate capacitors are connected in series to a battery of 100 $V$ velocity of the fluid at the point of measurement . 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
Text Solution
Verified by ExpertsThe correct answer is:
B
$c_{eq} = \frac{c \times 4c}{(c + 4c)} = \frac{4c}{5}$
$Q = C_{eq} \cdot V = \frac{4C}{5} \times 100 = 80C$
Hence $V_1 = \frac{Q}{C_1} = \frac{80\,C}{C_1} = 80\,V$
and $V_2 = \frac{80c}{4c} = 20\,V$
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