Question

Separation between the plates of a parallel plate capacitor is 5 mm. This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3 mm and dielectric constant K = 10 is placed between the plates, as shown.

Potential difference between the plates after the dielectric slab has been introduced is –

• A. 18.5 V
• B. 13.5 V
• C. 11.5 V
• D. 6.5 V

Answer 1

Answer: (C)

11.5 V

Answer 2

The capacitor is charged by a battery of 25 V. Let the magnitude of surface charge density on each plate be σ. Before inserting the dielectric slab, electric field strength between the plates,

E = $\frac { \sigma } { \varepsilon _ { 0 } }$=

or E = $\frac { \sigma } { \varepsilon _ { 0 } }$ = $\frac { 25 } { 5 \times 10 ^ { - 3 } }$ = 5000 N/C

The capacitor is disconnected from the battery but charge on it will not change so that σ has the same value. When a dielectric slab of thickness 3mm is placed between the plates, the thickness of air between the plates will be 5 – 3 = 2 mm. Electric field strength in air will have the same value (5000 N/C) but inside the dielectric, it will be = $\frac { 5000 } { 10 }$

= 500 N/C

so potential difference = E_air d_air + E_med d_med

= 5000 × (2 × 10^–3) + 500 × (3 × 10^–3)

= 11.5 V