Physics, asked by anirudhmovva16, 3 months ago

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

capacitor plate

Answers

Answered by iitianairfor2026
0

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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