**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

**A.** the charges on plates 3 and 6 are +Q and -Q

Correct answeris C.+Q and -Q

**B.** When capacitors are connected in series, the charge on each is same and equal to total charge stored

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

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

V'=Q /C

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

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

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

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

**D.** The charge Q on the first capacitorQ = 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

**E.** The equivalent capacitance C_eq for this combination ofcapacitors in series.

C_eq =6c /11