Question

23. A parallel-plate capacitor is placed in such a way that
its plates are horizontal and the lower plate is dipped
into a liquid of dielectric constant K and density p. Each
plate has an area A. The plates are now connected to a
battery which supplies a positive charge of magnitude Q
to the upper plate. Find the rise in the level of the liquid
in the space between the plates.
tc-Q1 - 1)
QK​

Answer 1

Answer:

Taking the electric filed above the liquid as  E1 = Q/(εoA)

Taking the electric field below the liquid as E2 = Q/(εoK A)

Hence, the net electric field will be.

E2 - E1 = Q/(εoA) - Q/(εoK A).

 E = Q(1 – 1/K)/(εoA).

Since, the charge induced on the surface is q.

As the angle will be 180 since they lie on same plane.

{Q(1 – 1/K)/(εoA)}Acos(180) = q/εo.

On solving we get.

q = -Q(1 – 1/K).

Hence at bottom the charge will be  –Q – { -Q(1 – 1/K)} = -Q/K.

Now on the surface we know E1 = Q/(2εoA).

Forces will be.

F1 = qE1.

F1 = Q2(1 – 1/K)/(2εoA).

E2 = Q/(2εoK).

F2 =  qQ/(2εoK).

F2 = Q2(1 – 1/K)/(2εoAK).

Hence the net force will be.

F = F1 + F2

.

Q2(1 – 1/K)/(2εoAK) + Q2(1 – 1/K)/(2εoA).

Q2(1 – 1/K) ( 1 + 1/K)/(2εoA).

Q2( 1 – 1/K2)/2εoA.

Q2( K2 – 1)/(2K2εoA).

F = ρAhg.  

Q2( K2 – 1)/(2K2εoA) = ρAhg.

h =  Q2( K2 – 1)/(2ρgK2εoA2).