Verified answer

Easy,

(a) if parallel, add all capacitance...

Total C = C1 C2

C = 2.20 3.60 = 5.8 µF...

given Q = CV

Q = (5.8 x 10^-6) x 57.6

Q = 3.3408 x 10^-4

Q = 3.34 x 10^-4 coulombs (or 334.08µC)

.

(b) in series, add the reciprocal of all capacitance...

1/C = 1/C1 1/C2

1/C = 1/2.2 1/3.6

1/C = 0.7323

so, C = 1/0.7323 = 1.366 µF...

Q = CV

Q = 1.366 x 10^-6 x 57.6

Q = 7.87 x 10^-5 Coulombs.

First you need to work out the total capacitance in each case

(a) When in parallel, the two capacitors behave as a single capacitor of capacitance:

Cpar. = C1 + C2 = 2.20 + 3.60 = 5.80μF

The total charge stored in the two initial capacitors is therefore:

Qpar. = Cpar. X V

= 5.80 X 57.6

= 334.08μC, i.e. 334μC to 3s.f.

(b) When in series, the equivalent capacitance would be given by the relationship:

1/Cser. = 1/C1 + 1/C2

= 1/2.20 + 1/3.60

= 0.7323... => Cser. = 1.3655...μF

The total charge stored in the two capacitors in series would be:

Qser. = Cser. X V

= 1.3655... X 57.6

= 78.653...μC, i.e. 78.7μC to 3s.f.

**The final answer is given to an accuracy of 3s.f. because such is the (minimum) degree of accuracy in the given data.

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