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a parallel plate capacitor is charged to a potential difference of V by d.c. source.The capacitor is then disconnected from the source.If the distance between capacitor is doubled,state with reasons how the following will change: 1)electric field between the plates 2) Capacitance 3) Energy stored in capacitor

Grade:12

3 Answers

Apoorva Arora IIT Roorkee
askIITians Faculty 181 Points
9 years ago
The electric field is related to the potential as V= Ed.
so,E=\frac{V}{d}

when the distance is doubled,
E'=\frac{V}{2d}=\frac{E}{2}
So the electric field gets halved.

The capacitance is given by
C=\frac{A\epsilon _{0}}{d}
So, when the distance is doubled
C'=\frac{A\epsilon _{0}}{2d}=\frac{C}{2}
So the capacitance gets halved

The energy stored in a capacitor is given by
e=\frac{1}{2}CV^{2}
So, when the distance is doubled, the capacitance is halved and the potential is still the same.
e'=\frac{1}{2}\frac{C}{2}V^{2}=\frac{e}{2}
So, the energy also gets halved.

shabnam
15 Points
5 years ago
if the distance is doubled C is halved, V is doubled and Electric field remains same.
energy becomes twice 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

The electric field is related to the potential as V= Ed.
so, E =V/d when the distance is doubled,
E' = V/2d = E/2
So the electric field gets halved.
The capacitance is given by C = Aε/d
So, when the distance is doubled
C' =Aε/2d = C/2
So the capacitance gets halved
The energy stored in a capacitor is given by e = 0.5CV^{2}
So, when the distance is doubled, the capacitance is halved and the potential is still the same.
e' = 0.5*(C/2)*V^2
= e/2
So, the energy also gets halved.

Thanks and Regards

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