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

Answer:

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

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

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

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}$

$V_{1}=9.18\ V$

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

$C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}$

$C=6.99\times10^{-7}\ F$

$C=0.69\ \mu F$

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

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