A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another 300 pF capacitor. The electrostatic energy lost in the process is

Question

A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another 300 pF capacitor. The electrostatic energy lost in the process is (A) zero (B)  4× 10-6J  (C)  6× 10-6J  (D)  8× 10-6J

Tardigrade Admin · Accepted Answer

Energy stored in capacitor E1=(1/2) C V2 =(1/2) × 600 × 10-12 ×(200)2 =12 × 10-6 J Again E2=(1/2)(600+300) × 10-12 ×(200)2 =(1/2) × 900 × 10-12 × 4 × 104 =18 × 106 J Energy lost =E1=-E2 =18 × 10-6-12 × 10-6 =6 × 10-6 J

Q. A $600 pF$ capacitor is charged by $200 V$ supply. It is then disconnected from the supply and is connected to another $300 pF$ capacitor. The electrostatic energy lost in the process is

zero

$4 \times 1 0^{- 6} J$

$6 \times 1 0^{- 6} J$

$8 \times 1 0^{- 6} J$

Q. A 600pF capacitor is charged by 200V supply. It is then disconnected from the supply and is connected to another 300pF capacitor. The electrostatic energy lost in the process is

zero

4×10−6J

6×10−6J

8×10−6J

Energy stored in capacitor E1​=21​CV2 =21​×600×10−12×(200)2 =12×10−6J Again E2​=21​(600+300)×10−12×(200)2 =21​×900×10−12×4×104 =18×106J Energy lost =E1​=−E2​ =18×10−6−12×10−6 =6×10−6J

Q. A $600 pF$ capacitor is charged by $200 V$ supply. It is then disconnected from the supply and is connected to another $300 pF$ capacitor. The electrostatic energy lost in the process is

$4 \times 1 0^{- 6} J$

$6 \times 1 0^{- 6} J$

$8 \times 1 0^{- 6} J$