On increasing the plate separation of charged condenser its energy :
decreases
remains unchanged
increases
none of these

Question

On increasing the plate separation of charged condenser its energy -decreasesremains unchangedincreasesnone of these

Toppr Admin · Accepted Answer

Parallel plate capacitance - C-A-x3F5-0dPotential between the plate is V-QdA-x3F5-0Energy U-12CV2-12-xD7-A-x3F5-0d-xD7-QdA-x3F5-0-2thus- U-x221D-dSo when the separation d increases- energy U will increase-

On increasing the plate separation of charged condenser its energy :decreasesremains unchangedincreasesnone of these

Parallel plate capacitance , C=Aϵ0dPotential between the plate is V=QdAϵ0Energy U=12CV2=12×Aϵ0d×(QdAϵ0)2thus, U∝dSo when the separation d increases, energy U will increase.

On increasing the plate separation of charged condenser its energy :
decreases
remains unchanged
increases
none of these

Parallel plate capacitance , $C = \frac{A ϵ_{0}}{d}$
Potential between the plate is $V = \frac{Q d}{A ϵ_{0}}$
Energy $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times \frac{A ϵ_{0}}{d} \times (\frac{Q d}{A ϵ_{0}})^{2}$
thus, $U \propto d$
So when the separation d increases, energy U will increase.