Question

The capacitance of parallel plate capacitor is 50pF and the distance between the plates is 4 mm. It is charged to 200 V, after that the chargnig battery is removed. Now, a dielectric slab of thickness 2 mm and dielectric constant 4 is placed between the plates. The exess heat produced in the slab, during this process, is;

(1) 1.0 x 10^-6 J

(2) 6.25 x 10^-7 J

(3) 3.75 x 10^-7 J

(4) 2.50 x 10^-7 J

Answer 1

(3) 3.75 x 10^-7 J

With air as dielectric, the capacitance of a parallel plate capacitor is;

Now, with a dielectric slab of thickness t and dielectric constant K, the capacitance is:

When C₀ is connected to the battery, then

After charging the battery is removed. Hence the change Q will remain unchanged, there after. How, suppose the capacitance changes to C, after placing the dielectric.

Now, energy loss, when the slab is placed

This energy loss will appear as surplus heat produced in the slab. So heat produced in the slab = 3.75 x 10^-7 J