A potential difference of 200 V is applied to a series connection of two capacit

Question

A potential difference of 200 V is applied to a series connection of two capacitors, of capacitances C1 = 3.00 µF and capacitance C2 = 8.00 µF.

(a) What is the charge q1 on capacitor 1?
C

(b) What is the potential difference V1 across capacitor 1?
V

(c) What is the charge q2 on capacitor 2?
C

(d) What is the potential difference V2 on capacitor 2?
V
The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used).
(e) What now is q1?
C

(f) What is V1?
V

(g) What is q2?
C

(h) What is V2?
V
Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together.
(i) What now is q1?
C

(j) What is V1?
V

(k) What is q2?
C

(l) What is V2?
V

Explanation / Answer

The potential difference V = 200V the capacitors C1 = 3.0F and C2 = 8.0F When there are connect in series     1/C = 1/C1 + 1/C2    Ceq = C1 C2 / C1 + C2           = (3.0*10^-6)(8.0*10^-6) / (3.0+8.0)*10^-6           = 2.2*10^-6 F Then the charge net charge        Q = Ceq V = (2.2*10^-6)(200)              = 440*10^-6 C (a) the charge on capacitor C1 is           Q1 = 440*10^-6C (b) The potential across capacitor C1 is             V1 = Q1/C 1                   = (440*10^-6) / 3.0*10^-6F)                   = 146. V (c) when they connected in series the charge on each is same so Q2 = 436*10^-6 C (d) The potential difference V2 = Q2/C2                                                 = (440*10^-6) / (8.0*10^-6)                                                 = 55 V

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