Question

In the figure (Figure 1) , each capacitor has 5.00 μF and Vab = 35.0 V .

a) Calculate the charge on each capacitor(Q1,Q2,Q3,Q4)

b) Calculate the potential difference across each capacitor

c) calculate the potential difference between points a and d

Answer 1

C1 and C2 are in series and are equivalent to C12

$\frac{1}{C_{12}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

C12 and C3 are in parallel and are equivalent to C123

C123 = C12 + C3 = 2.5 * 10^-6 + 5 * 10^-6 = 7.5 * 10^-6 F

C123 and C4 are in series and are equivalent to C1234

$\frac{1}{C_{1234}} = \frac{1}{C_{123}} + \frac{1}{C_{4}}$

$Q_{1234} = C_{1234} * V = 3 * 10^{-6} * 35 = 105 \ \mu C$

charge is same for capacitors in series

$Q_{123} = Q_4 = Q_{1234} = 105 \ \mu C$

Then

$V_{123} = \frac{Q_{123 }}{C_{123 } } = 14 V$

$V_{4} = \frac{Q_{4}}{C_{4} } = 21 \ V$

Now

, as it should

Next considering the circuit C12 , C3 and C4

$Q_3 = 70 \ \mu C$

$Q_{12} = C_{12} V_{12}= 35 \ \mu C$

Now consider the original Circuit

$Q_1 = Q_2 = Q_{12} = 35 \ \mu C$ , as charge is same for capacitors in series

$V_1 = \frac{Q_1}{C_1} = 7 \ V$

$V_2 = \frac{Q_2}{C_2} = 7 \ V$

Now V1 + V2 = 14 V , which equals V3 , as it should

Summary :

$Q_1 = 35 \mu C\ \ , \ \ V_1 = 7 V$

$Q_2= 35 \mu C\ \ , \ \ V_2 = 7 V$

$Q_3 = 70 \ \mu C\ \ \ \ , \ \ V_3 = 14 \ V$

$Q_4 = 105 \ \mu C \ \ , \ \ V_4 = 21$

c)Calculate the potential difference between points a and d

V_ad = V3 = 14 V