Question

When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?

Answer 1

Between the capacitor’s plates, when the dielectric slab is increased, the plates’ charge stays constant. This is because the plates are inaccessible.  

It is observed that the plates area does not change. So, the force remains constant between the plates.

EXPLANATION:

Between the capacitor of the plates, force is given by,  

$F=\frac{Q^{2}}{2 A \epsilon_{0}}$

• where, A is the area of plates, Q is the capacitor’s charge.  
• Also, F=QE

• where, E is the intensity of electric field and Q is the point charge’s test charge.

On one plate, field due to charge Q is  

$E=\frac{Q}{2 A \epsilon_{0}}$

So,  

$F=\frac{Q(-Q)}{2 A \epsilon_{0}}=-\frac{Q^{2}}{2 A \varepsilon_{o}}$

So, if the force among the plates stay constant, the plates area does not change.