Question

Figure (31-E30) shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l₁ and l₂. The left half of the dielectric slab has a dielectric constant K₁ and the right half K₂. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.

Answer 1

(a) Consider the left side

The plate area of the part with the dielectric is by its capacitance

Suppose dielectric slab is attracted by electric field and an external force F consider the part dx which makes inside further, As the potential difference remains constant at V.

The charge supply, dq = (dc)v to the capacitor

The work done by the battery is dw_b = v.dq = (dc)v²

The external force F does a work dwe = (–f.dx) during a small displacement

The total work done in the capacitor is dw_b + dw_e = (dc)v² – fdx

This should be equal to the increase dv in the stored energy.

∴ The ratio of the emf of the left battery to the right battery