A parallel plate capacitor has a capacitance of 19.2 μ F, when the insulator between the plates is glass, K=8.0. When the glass is replaced with rubber, the capacitance between the plates is 6.0 μ F. What is the dielectric constant of the rubber ?

Question

A parallel plate capacitor has a capacitance of 19.2 μ F, when the insulator between the plates is glass, K=8.0. When the glass is replaced with rubber, the capacitance between the plates is 6.0 μ F. What is the dielectric constant of the rubber ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(C1=(ϵ0 K1 A/d) = 19.2/C2=(ϵ0 K A)d = 6) or (8/K) =(19.2/6) K =(48/19.2)=2.5

Q. A parallel plate capacitor has a capacitance of $19.2 μ F$ , when the insulator between the plates is glass, $K = 8.0$ . When the glass is replaced with rubber, the capacitance between the plates is $6.0 μ F$ . What is the dielectric constant of the rubber ?

$\frac{C _{1} = \frac{ϵ _{0} K _{1} A}{d} = 19.2}{C _{2} = \frac{ϵ _{0} K A}{d} = 6}$
or $\frac{8}{K} = \frac{19.2}{6}$
$K = \frac{48}{19.2} = 2.5$

Q. A parallel plate capacitor has a capacitance of 19.2μF, when the insulator between the plates is glass, K=8.0. When the glass is replaced with rubber, the capacitance between the plates is 6.0μF. What is the dielectric constant of the rubber ?

C2​=dϵ0​KA​=6C1​=dϵ0​K1​A​=19.2​ or K8​=619.2​ K=19.248​=2.5

Q. A parallel plate capacitor has a capacitance of $19.2 μ F$ , when the insulator between the plates is glass, $K = 8.0$ . When the glass is replaced with rubber, the capacitance between the plates is $6.0 μ F$ . What is the dielectric constant of the rubber ?

$\frac{C _{1} = \frac{ϵ _{0} K _{1} A}{d} = 19.2}{C _{2} = \frac{ϵ _{0} K A}{d} = 6}$
or $\frac{8}{K} = \frac{19.2}{6}$
$K = \frac{48}{19.2} = 2.5$