Question

If q_f  is the free charge on the capacitor plates and q_b  is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge q_b can be expressed as:

Answer 1

Correct option is (2)

When a dielectric is inserted in a capacitor

Due to free charge \(\vec E=\vec E_0\) only

After dielectric \(E'=\cfrac{E_0}k\)

q_B = q_f(1 - 1/k)