Question

a parallel-plate capacitor is oriented horizontally and charged so that there is an electric field of 50,000 V/m pointing up inside the gap of 1cm, which is filled with air. a.) With the charge on each plate remaining constant, the capacitor is immersed half-way in an oil of dielectric constant K=2.5 What is now the potential difference between the plates? b.) If on top of the oil, water (K=80) is poured to completely fill the upper part of the gap, what is now the potential difference between the plates? (since this measurement takes place over a short period of time, please treat water as an insulator, not as a conductor.)

Answer 1

a)

Capacitance of the capacitor, C = *A/d

Now, electric field inside, E = V/d = 50000

So, V = 50000*0.01 = 500 V

So, Charge on the capacitor initially , Q = C*V = (*A/d)*V

General expression of a partially filled capacitor is given by:

C = *A/(d + t*(1/k - 1))

where d = dispance between the plates

t =thickness of the dielectric = d/2

k = dielectric constant of oil = 2.5

So,

after introducing the oil, the charge on the capacitor,

Q = (*A/(d + t*(1/k - 1)))*V'

where V' = new voltage across the capacitor

Now,  (*A/(d + t*(1/k - 1)))*V' =  (*A/d)*V

So, (1/(d+(d/2)*(1/2.5 -1)))*V' = V/d

So, (1/(1 + (1/2.5-1)/2))*V' = V

So, V' = 500/(1/(1 + (1/2.5-1)/2)) = 350 V

b)

Now, after filling water,

C =   *A/((d/2)/k + (d/2)/k')

where k' = 80

So, (*A/((d/2)/k + (d/2)/k'))*V' = (*A/d)*V

So, (2/(1/k+1/k'))*V' = V

So, V' = (500/2)*(1/80+1/2.5) = 103.1 V <-------answer