Question

if the potential of a capacitor having capacity 8pF is increased from 10v to 20 v then increase in the energy will be​

Answer 1

Explanation:

Capacitance of capacitor (C) =6μF=6×10

−6

; Initial potential (V

1

)=10V and final potential (V

2

)=20V.

The increase in energy ΔU=

2

1

C(V

2

2

−V

1

2

)

=

2

1

×(6×10

−6

)×[(20)

2

−(10)

2

]

=(3×10

−6

)×300=9×10

−4

J