Question

A parallel plate capacitor is made of two plates of length l and width and separated by distance
d. A dielectric slab that fits exactly between the plates is held near the edge of the plates. It is pulled into the

Answer 1

Hey dear,
Incomplete question. I've attached the complete one in image.

● Answer-
F = -dkQ^2 / 2wεl^2

● Explaination-
Here, capacitors are arranged in parallel combo.
C = C1 + C2
C = K(x)wε/d + (l-x)wε/d
C = wε/d (kx+l-x)

Energy of the capacitor is given by,
U = Q^2 / 2C
U = dQ^2 / 2wε(kx+l-x)

Force on dielectric,
F = -dU/dx
F = -dQ^2(k-1) / 2wε
[l+(k-1)x]^2
F = -dkQ^2 / 2wεl^2

Hope this is useful...