The capacitor of an oscillatory circuit of negligible resistance is enclosed in a container. When the container is evacuated, the frequency of the circuit is `150 kc//s` and when the container is filled with a gas, the frequency changes by `100 C//s`. Find the dielectric constant of the gas.

Explanation

Here, `v = 150 kc//s = 1.5 xx 10^(5) C//s`
`v' = v - 100 = (1.5 xx 10^(5) - 100) C//s`
`= 149900 C//s`
The frequency of oscillation in vacuum is
`v = (1)/(2 pi sqrt(LC))`
On introducing the gas. Capacity of condenser become K time. As K gt 1.
Capacity increases, The resonant frequency decreases by `100 C//s` Now, `C' = KC`
`v = (1)/(2 pi sqrt(LC')) = (1)/(2 pi sqrt(LKC))`
Dividing, we get
`(v)/(v') = sqrt(K)`
or `K = ((v)/(v'))^(2) = ((1.5 xx 10^(5))/(149900))^(2)`
` = ((1.5)/(1.499))^(2) = (1 + (0.001)/(1.499))^(2)`
`K = (1 + (0.002)/(1.499)) = 1.0013`