Question

a)show that the potential energy on a capacitor is U=Q^2/(2c) for a capacitor C holding charge Q.

b)what is the total capaciatnce of 5 eqqual capacitors C connected in paralledl? in series?

c)Two parallel plate capacitors are connected in series.Both are filled with a dielectric with constant K.One capacitor had plates of area A1 seperated by distance d1, the other has plate of area A2 seperated by distance d2.What is the toal capacitance of the two connected in series.

Answer 1

a) DU = q DV,

and the potential V increases as the charge is placed on the plates (V = Q / C). Since the V changes as the Q is increased, we have to integrate over all the little dq being added to a plate:

DU = q DV leads to

U = int( V dq ) where int is integration.

= int( q/c dq)

= 1/C int( q dq)

= Q^2 / 2C.

b)Parallel: 5C

Series : C/5.

c)C1 = Ke0A1/d1

C2 = Ke0A2/d2

C1 & C2 in series

Ceq = C1*C2/(C1+C2)

= (Ke0A1/d1)*(Ke0A2/d2)/{Ke0A1/d1 + Ke0A2/d2}

=Ke0*A1*A2/(A1d2 + A2d1)