Showthat the force on each plate of a parallel plate capacitor has amagnitude equal to (½) QE,where Q isthe charge on the capacitor, and E isthe magnitude of electric field between the plates. Explain theorigin of the factor ½.
Showthat the force on each plate of a parallel plate capacitor has amagnitude equal to (½) QE,where Q isthe charge on the capacitor, and E isthe magnitude of electric field between the plates. Explain theorigin of the factor ½.
Let F be theforce applied to separate the plates of a parallel plate capacitor bya distance of Δx.Hence, work done by the force to do so = FΔx
As a result, thepotential energy of the capacitor increases by an amount given asuAΔx.
Where,
u = Energydensity
A = Area of eachplate
d = Distancebetween the plates
V = Potentialdifference across the plates
The work done will beequal to the increase in the potential energy i.e.,
Electric intensity isgiven by,
However, capacitance, 
Charge on the capacitoris given by,
Q = CV
The physical origin ofthe factor, 
,in the force formula lies in the fact that just outside theconductor, field is E and inside it is zero. Hence, it is theaverage value, 
,of the field that contributes to the force.
Let F be theforce applied to separate the plates of a parallel plate capacitor bya distance of Δx.Hence, work done by the force to do so = FΔx
As a result, thepotential energy of the capacitor increases by an amount given asuAΔx.
Where,
u = Energydensity
A = Area of eachplate
d = Distancebetween the plates
V = Potentialdifference across the plates
The work done will beequal to the increase in the potential energy i.e.,
Electric intensity isgiven by,
However, capacitance,
Charge on the capacitoris given by,
Q = CV
The physical origin ofthe factor, ,in the force formula lies in the fact that just outside theconductor, field is E and inside it is zero. Hence, it is theaverage value, ,of the field that contributes to the force.
