An air filled capacitor of capacitance 9.7 pF constituted by 2 identical aluminum sheets separated by 1.20 mm is charged to 13.0 V. Calculate the area of sheets used. Keeping the charge held by the system constant, the plate separation is decreased by 0.10 mm. Find the new capacitance and potential difference between the 2 sheets.
An air filled capacitor of capacitance 9.7 pF constituted by 2 identical aluminum sheets separated by 1.20 mm is charged to 13.0 V. Calculate the area of sheets used. Keeping the charge held by the system constant, the plate separation is decreased by 0.10 mm. Find the new capacitance and potential difference between the 2 sheets.
Given –
C = 9.7 pF = 9.7*10-12 F
d = 1.20 mm = 1.20*10-3 m
Potential difference = V = 13.0 V
Therefore, area of capacitor, A = Cd/ ε0 = (9.7*10-12*1.2*10-3) / 8.885*10-12
Or, A = 1.315*10-3 m2.
New separation, d’ = 1.20-1.10 = 1.10 mm = 1.10*10-3 m.
Therefore, new capacitance = C’ = A ε0/ d = (1.315*10-3*8.85*10-12) / 1.1*10-3 = 10.58*10-12 F
New capacitance = C’= 10.58 pF.
New potential difference, v’ = vC/C’ = (13*9.7*10-12) /10.58*10-12 = 11.920 V.
New potential difference, V’ = 11.920 V.
Given –
C = 9.7 pF = 9.7*10-12 F
d = 1.20 mm = 1.20*10-3 m
Potential difference = V = 13.0 V
Therefore, area of capacitor, A = Cd/ ε0 = (9.7*10-12*1.2*10-3) / 8.885*10-12
Or, A = 1.315*10-3 m2.
New separation, d’ = 1.20-1.10 = 1.10 mm = 1.10*10-3 m.
Therefore, new capacitance = C’ = A ε0/ d = (1.315*10-3*8.85*10-12) / 1.1*10-3 = 10.58*10-12 F
New capacitance = C’= 10.58 pF.
New potential difference, v’ = vC/C’ = (13*9.7*10-12) /10.58*10-12 = 11.920 V.
New potential difference, V’ = 11.920 V.
