A parallel plate capacitor of capacitance 4uf is charged using a 200V battery. After the capacitor charged the battery is removed and the distance between the capacitor plates is doubled. Calculate the new energy in the capacitor.
Answers
Answer:
Capacitance of a charged capacitor, C
1
=4μF=4×10
−6
F
Supply voltage, V
1
=200V
Electrostatic energy stored in C1 is given by,
E
1
=
2
1
C
1
V
1
2
=
2
1
×4×10
−6
×(200)
2
=8×10
−2
J
Capacitance of an uncharged capacitor, C
2
=2μF=2×10
−6
F
When C
2
is connected to the circuit, the potential acquired by it is V
2
.
According to the conservation of charge, initial charge on capacitor C
1
is equal to the final charge on capacitors, C
1
and C
2
∴V
2
(C
1
+C
2
)=C
1
V
1
V
2
×(4+2)×10
−6
=4×10
−6
×200
V
2
=
3
400
V
Electrostatic energy for the combination of two capacitors is given by,
E
2
=
2
1
(C
1
+C
2
)V
2
2
=
2
1
(2+4)×10
−6
×(
3
400
)
2
=5.33×10
−2
J
Hence, amount of electrostatic energy lost by capacitor C
1
=E
1
−E
2
=0.08−0.0533=0.0267
=2.67×10
−2
J