A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

Question

A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ QL = (Q2/2C)-[((Q/ 2)2/2C)×2] = (Q2/4C) = (1/4)CV2 = (1/4)Ã 60 Ã 10-12 Ã 4 Ã 102 = 6nJ

Q. A $60 pF$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

$Δ Q_{L} = \frac{Q ^{2}}{2 C} - [\frac{( Q /2 ) ^{2}}{2 C} \times 2] = \frac{Q ^{2}}{4 C}$
$= \frac{1}{4} C V^{2}$
$= \frac{1}{4} \times 60 \times 1 0^{- 12} \times 4 \times 1 0^{2}$
$= 6 n J$

Q. A 60pF capacitor is fully charged by a 20V supply. It is then disconnected from the supply and is connected to another uncharged 60pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

ΔQL​=2CQ2​−[2C(Q/2)2​×2]=4CQ2​ =41​CV2 =41​×60×10−12×4×102 =6nJ

Q. A $60 pF$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

$Δ Q_{L} = \frac{Q ^{2}}{2 C} - [\frac{( Q /2 ) ^{2}}{2 C} \times 2] = \frac{Q ^{2}}{4 C}$
$= \frac{1}{4} C V^{2}$
$= \frac{1}{4} \times 60 \times 1 0^{- 12} \times 4 \times 1 0^{2}$
$= 6 n J$