Question

A.) What are the charges on plates 3 and 6?

B.) If the voltage across the first capacitor (the one with capacitance C ) is \({V^{\prime}}\), then what are the voltages across the second and third capacitors?

C.) Find the voltage \(V_{1}\) across the first capacitor.

D.) Find the charge Q on the first capacitor.

E.) Using the value of Q just calculated, find the equivalent capacitance \(C_{\mathrm{eq}}\) for this combination of capacitors in series.

Answer 1

1) the charges on plates 3 and 6 are +Q and -Q

The correct answer is C.+Q and -Q

 

2) When capacitors are connected in series, the charge on each is the same and equal to the total charge stored

In series combination, the potential across any capacitor is inversely proportional to its capacity

If the voltage across the first capacitor (the one with capacitance C) is V', then

V'=Q /C

The voltage across the second capacitor =Q/2C =(1/2)Q/C=V'/2

The voltage across the third capacitor =Q/3C =(1/3)Q/C=V'/3

The correct answer is B.V'/2 and V'/3

V= V_1 +V_2+V_3

V= V_1 +V_1/2+V_1/3

V = 11V_1/6

 

3 .The voltage V_1 across the first capacitor= V_1= 6V/ 11

 

4 The charge Q on the first capacitor  = cV_1

V= V_1 +V_2+V_3

Q/C_eq =Q/c +Q/2c +Q/3c

1/C_eq =1/c +1/2c +1/3c

 

5 The equivalent capacitance C_eq for this combination of capacitors in series.

C_eq =6c /11