A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is nJ.

Question

A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is nJ. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy loss =(1/2) (C1 C2/C1+C2)(V1-V2)2 =(1/2) (50 × 50 × 10-12 × 10-12/(50+50) 10-12)(100-0)2=125 n J

Q. A capacitor of capacitance $50 pF$ is charged by $100 V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _______ $n J$ .

Energy loss $= \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$
$= \frac{1}{2} \frac{50 \times 50 \times 1 0 ^{- 12} \times 1 0 ^{- 12}}{( 50 + 50 ) 1 0 ^{- 12}} (100 - 0)^{2} = 125 n J$

Q. A capacitor of capacitance 50pF is charged by 100V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _______nJ.

Energy loss =21​C1​+C2​C1​C2​​(V1​−V2​)2 =21​(50+50)10−1250×50×10−12×10−12​(100−0)2=125nJ

Q. A capacitor of capacitance $50 pF$ is charged by $100 V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _______ $n J$ .

Energy loss $= \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$
$= \frac{1}{2} \frac{50 \times 50 \times 1 0 ^{- 12} \times 1 0 ^{- 12}}{( 50 + 50 ) 1 0 ^{- 12}} (100 - 0)^{2} = 125 n J$