201.A capacitor of capacitance
1/300 microfarad
connected to a battery of 300 V and
charged. Then the energy stored in the
condenser is
Answers
Answer:
q
o
=300×200×10
−12
=6×10
−8
C
E
o
=
2
1
CV
2
=
2
1
×200×10
−12
×(300)
2
=9×10
−6
J
Capacitor 1 will charge capacitor 2 until V
AB
=V
CD
200×10
−12
q
1
=
100×10
−12
q
2
or, q
1
=2q
2
But, q
1
+q
2
=q
o
=6×10
−8
∴3q
2
=6×10
−8
q
2
=2×10
−8
q
1
=4×10
−8
Energy stored in capacitor 1
=
2C
1
(q
1
)
2
=
2×200×10
−12
(4×10
−8
)
2
=4×10
−6
J
Energy stored in capacitor 2
=
2C
2
(q
2
)
2
=
2×200×10
−12
(2×10
−8
)
2
=2×10
−6
J
Difference in energy stored
(9×10
−6
J)−[4×10
−6
+2×10
−6
]J
=3×10
−6
J
solution
Explanation:
qo =300×200×10 −12=6×10−8C
EE=21
CV2= 21 ×200×10 −12 ×(300) 2=9×10 −6 J
Capacitor 1 will charge capacitor 2 until V
AB =V CD200×10 −12
q1=100×10 −12
q2or, qq =2q 2
But, q1+q2=qo=6×10 −8
∴3q2 =6×10−8
q2 =2×10−8
q1 =4×10−8
Energy stored in capacitor 1 = 2C (q 1 ) 2×200×10-12(4×10 −8 ) 2=4×10 −6 J
Energy stored in capacitor 2 = 2C 2(q 2)2×200×10−12(2×10 −8
)
2
=2×10
−6
J
Difference in energy stored
(9×10
−6
J)−[4×10
−6
+2×10
−6
]J
=3×10
−6
J