Question

A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are the charge and the potential difference for each capacitor? (b) The charged capacitors are reconnected with their positive plates together and their negative plates together, no external voltage being applied. What are the charge and the potential difference for each? (c) The charged capacitors in (a) are reconnected with plates of opposite sign together. What are the charge and the potential difference for each?

Answer 1

Answer:

a) Q1=Q2=480μC   V1=240V   V2=60V

b) Q1=96μC   Q2=384μC   V1=V2=48V

c) Q1=Q2=0C    V1=V2=0V

Explanation:

Let C1 = 2μC  and  C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

$C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C$

$Q_{T} = V_{T}*C_{T} = 480\mu C$

$V1 = \frac{Q1}{C1}=240V$

$V2 = \frac{Q2}{C2}=60V$

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

$V1=V2$   So,  $\frac{Q1}{C1} = \frac{Q2}{C2}$

Total charge is the same calculated for part (a), so:

$\frac{Qt - Q2}{C1} = \frac{Q2}{C2}$   Solving for Q2:

Q2 = 384μC    Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C  V1=V2=0V