A 200pF capacitor is connected to a 100 V battery. What is the charge on the
capacitor plate
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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
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