(b) The capacitor in a is now disconnected from the supply and connected to an uncharged 470 μF capacitor.
(i) Explain why the capacitors are in parallel, rather than series, and why the total charge stored by the combination must be the same as the answer to (a).
(ii) Calculate the capacitance of the combination.
(iii) Calculate the potential difference across each capacitor.
(iv) Calculate the charge stored on one plate of each capacitor.
Solution:
a)
Q = CV
= (470 x 10^-6) x (20)
= 9.4 mC
bi)
When capacitors are connected in series, the total capacitance is less than any one of the series capacitors’ individual capacitances. If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. As we’ve just seen, an increase in plate spacing, with all other factors unchanged, results in decreased capacitance.
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors’ capacitances. If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors. As we’ve just seen, an increase in plate area, with all other factors unchanged, results in increased capacitance.
The total charge stored by the combination must be the same as the answer to (a) is because the total charge of a capacitor is equal to the total charge of both capacitors.
bii)
C = C1 + C2
= 470 uF + 470 uF
= 940 uF
biii)
Q ∝ V1Q = 20V
0.5Q = 10V
biv)
Q1 = (C1) (V)
= (470 uF)(10)
= 4.7 mC
Q2 = (C1) (V)
= (470 uF)(10)
= 4.7 mC
Reference:
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