NCERT Solution : Electrostatic Potential & Capacitance

Q.5. A parallel plate capacitor with air between the plates has a
capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?

Sol. C = εoA/d =  8 pF

C’ = εoKA/d = K(εoA/d) = 6 x 8 = 48 pF

Q.6. Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination ?
(b) What is the potential difference across each capacitor if the
combination is connected to a 120 V supply ?

Sol. C = 9 pF

(a) As Three capacitors are in series

1/C’ = 1/C + 1/C + 1/C

1/C’ = 1/9 + 1/9 + 1/9

On solving , C’ = 3 pF

(b) Potential difference across each capacitor is ,

V = 120/3 = 40 volt

Q.7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is
connected to a 100 V supply.

Sol. (a) Total capacitance , C = 2 + 3 + 4 = 9 pF

(b)As all capacitors are connected in parallel, hence P.D. across each is same .
V = 100 volt

q1 = C1 V = 2 × 100 = 200pC

q2 = C2 V = 3 × 100 = 300pC

q3 = C3 V = 4 × 100 = 400pC

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