Question

A fixed potential difference is applied between the plates of a parallel plate capacitor. When a dielectric slab of thickness 3 mm is placed between the plates after removing the battery the distance between the plates is to be increased by 2.4 mm to keep the potential difference between the plates same. Dielectric constant of the material of slab will be-

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (D)

5

Answer 2

V = $\frac { \mathrm { Q } } { \mathrm { C } _ { \mathrm { i } } }$ =

$\frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} }$ =

d = d + 2.4 – 3 + ̃ $\frac { 3 } { \mathrm { k } }$ = 0.6 ̃ k = $\frac { 30 } { 6 }$ = 5