A parallel plate air capacitor has a capacitance 18 μ F. If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes 72 μ F. The dielectric constant of the medium is

Question

A parallel plate air capacitor has a capacitance 18 μ F. If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes 72 μ F. The dielectric constant of the medium is (A) 4 (B) 9 (C) 12 (D) 2

Tardigrade Admin · Accepted Answer

C0=(ε0 A/d)=18 C=(K ε0 A/3 d)=72 On dividing Eq. (ii) by Eq. (i), we get (K/3)=(72/18)=4 ∴ Dielectric constant, K=12

Q. A parallel plate air capacitor has a capacitance $18 μ F$ . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes $72 μ F$ . The dielectric constant of the medium is

4

9

12

2

$C_{0} = \frac{ε _{0} A}{d} = 18$
$C = \frac{K ε _{0} A}{3 d} = 72$
On dividing Eq. (ii) by Eq. (i), we get
$\frac{K}{3} = \frac{72}{18} = 4$
$∴$ Dielectric constant, $K = 12$

Q. A parallel plate air capacitor has a capacitance 18μF. If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes 72μF. The dielectric constant of the medium is

4

9

12

2

C0​=dε0​A​=18 C=3dKε0​A​=72 On dividing Eq. (ii) by Eq. (i), we get 3K​=1872​=4 ∴ Dielectric constant, K=12

Q. A parallel plate air capacitor has a capacitance $18 μ F$ . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes $72 μ F$ . The dielectric constant of the medium is

$C_{0} = \frac{ε _{0} A}{d} = 18$
$C = \frac{K ε _{0} A}{3 d} = 72$
On dividing Eq. (ii) by Eq. (i), we get
$\frac{K}{3} = \frac{72}{18} = 4$
$∴$ Dielectric constant, $K = 12$