IIT JEE , AIEEE Forum - Physics , Electricity

bumbi.ganguly

derive an expression for energy in a capacitor?

astronautameya

force of attraction betn plates =(sigma square*A)/2 epsilon

energy is work done to bring them to 0 distance.

i.e (sigma square A *a *d)/2 epsilon *A

i.e. q^2/2c =1/2cv^2

sigma is charge density

pls rate if satisfied

chimanshu_007

hii ! ! ! ! ! !

Consider a capacitor of capacitence C and is charged at potential diff. V . The charge stored in the capacitor is given by

Q = CV

V = Q/C.....(1)

Let the capacitor is given a very small charge dQ by a battery at const. potential V , the amt of work done is given by

dW= (dQ)V

dW = Q/C .dQ

total work done in charging the capacitor is given by

_{[ ]}

^{[ ]} dW = _{[ 0]}^{[ Q]}Q/C.dQ = 1/C (Q²/2) FRM 0 TO Q

W = 1/2 Q²/ C

this work done in tyhe capacitor is in the form of potential energy

U = 1/2 . Q² / C

Q = CV

U = 1/2 C² V²/ C = 1/2CV²

C = Q/V

U = 1/2 Q.V²/ V

U = 1/2 QV

SO

U = 1/2 CV² = 1/2Q² /C = 1/2 QV

elessar_iitkgp

Very well explained chimanshu

chimanshu_007

here's the pic