Question : A potential difference is set up between the plates of a parallel plate capacitor by a battery and then the battery is removed if the distance between the plates is decreased then how the
(i) charge
(ii) potential difference
(iii) electric field
(iv) energy and
(v) energy density will change?
Doubt by Shaurya
Solution :
Battery is removed and distance (d) between the plates of the capacitor is decreased.
(i) Since the battery is removed so no new charge could be added and as per the law of conservation of changes, the total charge on the plates of the capacitor will remain same. Hence, the chrage will remain same.
(ii) We know
E=V/d
V=Ed
V=[σ/ε₀]d
V=(q/A)d/ε₀
V=qd/Aε₀
Clearly
V∝d
So when d decreses then potential also decreases.
(iii) We know
E=V/d
E=[qd/Aε₀]/d [From Part (iii)]
E=q/Aε₀
Clearly
Electric field is independ of d. So E will remain same.
(iv) We know
Energy
U=½qV
U=½q(qd/Aε₀) [From Part (ii)]
U=½q²d/Aε₀
Clarly
U∝d
So when d decreses then energy also decreases.
Energy Density (u)
u=½ε₀E²
As per part (iii) Electric Field remains constant.
So, Energy Density, reamins constant.
