Question

Plz answer fast........

Answer 1

Answer:

A charged Capacitor is a store of electrical potential energy.

To find the energy stored in a capacitor, let us consider a capacitor of capacitance C, with a potential difference V between the plates.

There is a charge +q on one plate and –q on the other.

Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.

Potential of capacitor =q/C.

Small amount of work done in giving an additional charge dq to the capacitor is

dW=q/C *dq

total work done in giving a charge Q to the capacitor is q.

Q=Q

W=1/C Q2/2

Energy stored in the capacitor

U=W=1/2 Q2/C

Put Q=CV

U=1/2 (CV)2/C =1/2 CV2

Put CV=Q

CV=QU=1/2 QV

CV=QU=1/2 QVThus, U=1/2 Q2/C =1/2 CV2=1/2 QV

U=1/2 Q2/C =1/2 CV2=1/2 QVWhen Q is in coulomb,V is in Volt ,C is in fared. Energy U is in joule.

Energy stored in a capacitor is in the form of electric field energy in the charged capacitor and it resides in the dielectric medium between the plates.

$\huge \boxed { \bf\fcolorbox{aqua}{yellow}{follow \: me}}$

Answer 2

A charged Capacitor is a store of electrical potential energy.

There is a charge +q on one plate and –q on the other.

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor is

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dq

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.Q=Q

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.Q=QW=1/C Q2/2

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.Q=QW=1/C Q2/2Energy stored in the capacitor

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.Q=QW=1/C Q2/2Energy stored in the capacitorU=W=1/2 Q2/C

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.Q=QW=1/C Q2/2Energy stored in the capacitorU=W=1/2 Q2/CPut Q=CV

There is a charge +q on one plate and –q on the other.Suppose the capacitor is charged gradually. At any stage ,the charge on the capacitor is q.Potential of capacitor =q/C.Small amount of work done in giving an additional charge dq to the capacitor isdW=q/C *dqtotal work done in giving a charge Q to the capacitor is q.Q=QW=1/C Q2/2Energy stored in the capacitorU=W=1/2 Q2/CPut Q=CVU=1/2 (CV)2/C =1/2 CV2

I hope this helps you....