Question

Suppose we increase the charge on a parallel plate capacitor from 28 C to 73 C and increase the separation distance of the plates from 11 mm to 47 mm.

the energy stored in the capacitor? (If Vo is the original stored energy,

What will be the factor that will increase the final stored energy will be #Uo. which one is that #?)

Answer 1

Energy stored in capacitor= (q^2)/(2C), where q is charge and C is capacitance.

Also, for parallel plate capacitors , C=A*o/d, where A is area, o is permittivity in free space and d is separation between the parallel plates.

So, energy stored= (q^2)/[(2A*o/d)] = (d*q^2)/(2A*o)

So,Initial energy = Uo = (11*28*28)/(2A*o)

Also, final energy=U1 = (47*73*73)/(2A*o)

So, U1/Uo = (47*73*73)/(11*28*28) = 29

So, factor of increase= 29.