- #1
bs vasanth
- 16
- 1
A parallel plate capacitor of circular cross section r>>d, and separation d. It is charged to potential V then disconnected from the circuit . What will be the work done in moving the capacitor from d to d1?
Answer:
Here the initial energy is 1/2CV^{2}
Where C=Aε/d
While moving the capacitor both the capacitance and potential change:
Cf= Aε/d1
Vf=∫E.dl (with the limits 0 to d1)
Vf= E*d1
We know that the E doesn't change ( Under the approx. as r>>d )
Therefore initial potential and final potential can be related as V= E*d
Vf=V*d1/d
Work done = 1/2( Cf*Vf^{2} - C*V^{2} )
= Aε/2 ((V*d1/d)^{2}/d1 - C*V^{2}/d )
= AεV^{2}( d1/d -1)/2d
I want to know if this answer is correct.
Answer:
Here the initial energy is 1/2CV^{2}
Where C=Aε/d
While moving the capacitor both the capacitance and potential change:
Cf= Aε/d1
Vf=∫E.dl (with the limits 0 to d1)
Vf= E*d1
We know that the E doesn't change ( Under the approx. as r>>d )
Therefore initial potential and final potential can be related as V= E*d
Vf=V*d1/d
Work done = 1/2( Cf*Vf^{2} - C*V^{2} )
= Aε/2 ((V*d1/d)^{2}/d1 - C*V^{2}/d )
= AεV^{2}( d1/d -1)/2d
I want to know if this answer is correct.
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