Question

A capacitor with plate separation d is charged to V volts. The battery is disconnected and a dielectric slab of thickness d/2 and dielectric constant 2 is inserted between the plates. The potential difference across the terminals becomes –

• A. V
• B. 2V
• C. $\frac{4V}{3}$
• D. $\frac{3V}{4}$

Answer 1

Answer: (D)

$\frac{3V}{4}$

Answer 2

V = =

V = =

= $\frac { \mathrm { q } } { \epsilon _ { 0 } \mathrm {~A} }$

= $\frac { \mathrm { qd } } { \epsilon _ { 0 } \mathrm {~A} }$ $\left( \frac { 1 } { 2 } + \frac { 1 } { 4 } \right)$

= $\frac { 3 } { 4 }$

= $\frac { 3 } { 4 }$ V