Question

A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one plate, which is at a potential of 51.5 V. (a) Find the potential on the negative plate of the 1.10 μF capacitor. (b) Find the equivalent capacitance of the two capacitors.

Answer 1

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Step-by-step explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

`V_(1)=(Q)/(C_(1))`

Put the value into the formula

`V_(1)=(10.1 *10^(-6))/(1.10*10^(-6))`

`V_(1)=9.18\ V`

The potential on the second plate

`V_(2)=V-V_(1)`

`V_(2)=51.5 -9.18`

`V_(2)=42.32\ v`

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

`C=(C_(1)*C_(2))/(C_(1)+C_(2))`

Put the value into the formula

`C=(1.10*10^(-6)*1.92*10^(-6))/((1.10+1.92)*10^(-6))`

`C=6.99*10^(-7)\ F`

`C=0.69\ \mu F`

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.