Q. 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
A) $50\ V,\ 50\ V$
B) $80\ V,\ 20\ V$
C) $20\ V,\ 80\ V$
D) $75\ V,\ 25\ V$
Explanation
(b) $C_{eq} = \frac{C \times 4C}{(C + 4C)} = \frac{4C}{5}$
$Q = C_{eq}.V = \frac{4C}{5} \times 100 = 80C$
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Hence $V_{1} = \frac{Q}{C_{1}} = \frac{80C}{C} = 80V$
$and\ V_{2} = \frac{80C}{4C} = 20V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.