JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Capacitance

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The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitor are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is:

(A) $\frac{V}{K + n}$
(B)V
(C) $\frac{( n + 1 ) V}{( K + n )}$ ✓ Correct
(D) $\frac{nV}{K + n}$

Answer: (C)

Solution

While the battery is connected, the two parallel capacitors

C

and

n C

both sit at

V

, holding a total charge

Q = C V + n C V = (n + 1) C V

. Once the battery is disconnected, this charge is trapped on the combination. Slipping a dielectric of constant

K

into the first capacitor raises its capacitance to

K C

, so the combined capacitance becomes

K C + n C = (K + n) C

while the charge is unchanged. The new common voltage is therefore

V^{'} = \frac{Q}{( K + n ) C} = \frac{( n + 1 ) V}{K + n}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Capacitance

Solution

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