Question

A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a slab of material with dielectric constant k = 2 is inserted between the plates. Which of the following statements is true?

1. The capacitance of the capacitor is doubled.
2. The voltage across the capacitor is doubled.
3. The charge on the plates is doubled.
4. The charge on the plates decreases by a factor of 2.
5. The electric field is doubled.

Answer 1

Capacitance of a parallel plate capacitor is given by C = ε_oA/d , when air is in between the plates

When dielectric is introduced , the new capacitance C’ = k ε_oA/d

Here k=2

∴ C’ = 2 ε_oA/d = 2C

This means capacitance is doubled

The charge Q on the [plate will remain the same

Now the potential V = Q/C

New potential V’ = Q/C’ = Q/2C = V/2

This means potential is halved

Again Electric field E = σ/ε_o

When dielectric is inserted E’ = σ/k ε_o = σ/2 ε_o =   E/2

Electric field is halved

So only the first statement is correct