A parallel-plate capacitor having plate are `100 cm ^2` and separation 1.0 mm holds a charge of `0.12 muC` when connected to a 120 V battery. Find the dielectric constant of the material filling the gap.

Explanation

The capacitance of the capacitor is `
`0.12muC)/(120V) = 1.0 X 10^(-9) F`
`. If Kbe the dielectric constant, the capacitance is also given by (Kepsion_0A)/(d). Thus, (Kx 8.85 X 10^(-12) Fm(-1) X 100 X 10(-4) m^2)/(1.0X10^(-3) m) = 1.0 X 10^(-9) F`
` or , K = 11.3