B.) If the voltage across the first capacitor (the one with capacitance $ C_{)} $ is $ V*[prime] $, what are the voltages across the second and third capacitors?

C.) Determine the voltage $ { }^{V_{1}} $ across the first capacitor.

(D.) Find the charge on the first capacitor $ [ ]*[ Q $.

Finding the equivalent capacitance for this series combination of capacitors using the value of $ Q $ you just calculated.

Answer
A. Charges on plates 3 and 6 are +Q and -Q

The correct answer is C.+Q and -Q

B. When capacitors are connected in series, their charges are the same and equal

Any capacitor has an inversly proportional potential across it when connected in series

The voltage across the first capacitor (the one with capacitance C) is V’.

V’=Q /C

Voltage across second capacitor =Q/2C =(1/2)Q/C=V’/2

The voltage across the third capacitor equals Q/3C + (1/3)Q/C = V’/3

This is the correct answer 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 across the first capacitor is V_1 = 6V/ 11

cV_1 is the charge on the first capacitor Q

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

This combination of capacitors in series has the equivalent capacitance C_eq. C_eq =6c /1