A capacitor of capacitance c is fully charged by connecting to a battery of emf E.It is disconnected from the battery.If the seperation between the plates of a capacitor is doubled,how will be the following change?

(a)Charge stored by the capacitor

(b)Potential difference across it?

(c)Field strength between the plates?

(d)Energy stored by the capacitor?

First always remember 1.when battery is remove after charging capacitor ,the charge remain constant while doing operation of capacitor (eg. Inserting dielectric)2. When battery remain connected and then we do operation then potential on capacitor is constant.A) when plates move the charge remain constant that is equal to =cvB)as plate distance double the capacitance decreases therefore potential increase asQ=cv Q constant as c decreases v increase.C)electeic feild remaon constant asE(initial)= v/d =q/(cd) (c=èa/d) therfore E =q/èaE(final)=Vnew/2d =q/(Cnew×2d) (Cnew=èa/2d) therfore E (final)=q/èaD) energy =1/2q^2/c as c decreases energy increases

Dear Student,
Please find below the solution to your problem.

First always remember 1.when battery is remove after charging capacitor ,the charge remain constant while doing operation of capacitor (eg. Inserting dielectric)2. When battery remain connected and then we do operation then potential on capacitor is constant.A) when plates move the charge remain constant that is equal to =cvB)as plate distance double the capacitance decreases therefore potential increase asQ=cv Q constant as c decreases v increase.C)electeic feild remaon constant asE(initial)= v/d =q/(cd) (c=èa/d) therfore E =q/èaE(final)=Vnew/2d =q/(Cnew×2d) (Cnew=èa/2d) therfore E (final)=q/èaD) energy =1/2q^2/c as c decreases energy increases

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