Question

Figure shows two identical capacitors C₁ and C₂ each of 1 µF capacitance connected to a battery of 6 V. Initially switch S is closed. After some time S is left open and dielectric slab of dielectric constant k = 3 are inserted to fill

completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted.

Answer 1

In C₁

charge qd = C dV = KCV

i.e. it increases K-times

Potential V remains the same as 6 V.

In C₂

charge qd = C dV remains constant

And Potential

Vd = \(\frac{V}{K}\)

i.e. it decreases K-times.