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