#### Question

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

**(a)** What is the total capacitance of the combination?

**(b)** Determine the charge on each capacitor if the combination is connected to a 100 V supply.

#### Solution

**(a) **Capacitances of the given capacitors are

`C_1=2pF`

`C_2=3pF`

`C_3=4pF`

For the parallel combination of the capacitors, equivalent capacitorC'is given by the algebraic sum,

C'=2+3+4=9 pF

Therefore, total capacitance of the combination is 9 pF.

**(b) **Supply voltage, *V* = 100 V

The voltage through all the three capacitors is same = *V* = 100 V

Charge on a capacitor of capacitance *C* and potential difference *V* is given by the relation,

*q* = *VC* … (i)

For C = 2 pF,

Charge = VC=100 x 2=200pC=`2xx10^-19C`

For C = 3 pF,

Charge = VC=100 x 3=300pC=`3xx10^-19C`

For C = 4 pF,

Charge = VC=100 x 4=400pC=`4xx10^-19C`