The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2μ F. the separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is :

Question

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2μ F. the separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is : (A) 15.6 μ F  (B) 11.2 μ F  (C) 19 μ F  (D) 22.4 μ F

Tardigrade Admin · Accepted Answer

C=(ε0 A/d) propto (1/d) Hence, (C1/C2)=(d2/d1) or (2/C2)=(0.2/0.4) or C2=(0.4 × 2/0.2)=4 μ F Final capacity is C=4 × k=4 × 28 =11.2 μ F

Q. The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 c m$ is $2 μ F .$ the separation is reduced to half and it is filled with a dielectric substance of value of $2.8.$ The final capacity of capacitor is :

$15.6 μ F$

$11.2 μ F$

$19 μ F$

$22.4 μ F$

$C = \frac{ε _{0} A}{d} \propto \frac{1}{d}$
Hence, $\frac{C _{1}}{C _{2}} = \frac{d _{2}}{d _{1}}$
or $\frac{2}{C _{2}} = \frac{0.2}{0.4}$
or $C_{2} = \frac{0.4 \times 2}{0.2} = 4 μ F$
Final capacity is
$C = 4 \times k = 4 \times 28$
$= 11.2 μ F$

Q. The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4cm is 2μF. the separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is :

15.6μF

11.2μF

19μF

22.4μF