A capacitor having a capacitance of `100 mu f` is changed to a potential difference of `50V.`(a) What is the magnitude of the charge on each plate? (b)The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted . Calculate the new potential difference between the plate .(c) What charge would have producted this potential difference in absence of the dielectric slab.(d) Find the charge induced at a surface of the dielectric slab.

Explanation

(a) Charge = capacitance xx potential diff.
` = 100 xx 10^(6) xx 50 = 5 m C `
(b) When dielectrical is introduced , potential difference decreases
` P. d . = Intial potential / Dielectric constant `
` = 50 / 2.5 = 20 V ` (c ) Charge = Capacitnace xx P.d . `
(d) `q' = q_i - q_f `
= ` 5 m C - 2 mC = 3mC.