Capacitanceof a capacitor is 100μF and potential difference between plates is 50volts thencharge stored on each plate is
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Answer:
Solution :
(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.
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Q = C V
Q = 100 x 10^-6 x 50
Q = 10^-4x 50
Q = 5 x 10^-3C
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