Question

A voltage has been applied across a capacitor. If the dielectric is replaced with another dielectric constant eight times as great and the voltage is reduced to half of what it was, the ENERGY STORED in the capacitor is how many times the original stored energy?

Answer 1

Capacitance C of a parallel palte capacitor is given by C = KeoA/d

where A = area = pi r^2, e0 = constnat = 8.85*10^-12,

d   = distance between the plates, K = dieelctric constant (=1 for air)

Chareg Q = CV where V = Volatge

Energy U = 0.5QV = 0.5 CV^2 = Q^2/2C

eletric field E = V/d

so now intiail eenrgy is Ui = 0.5 CV^2--------------------- 1

when given conditions applied, such as K = 8, V = 1/2

then enenrgy U = 0.5(8C)*(V^2/4)

so U = 2 * (Ui)

so energy stored will ne two times (twice) that of the intial eenrgy