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:

(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

[tex]V_{1}=\dfrac{Q}{C_{1}}[/tex]

Put the value into the formula

[tex]V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}[/tex]

[tex]V_{1}=9.18\ V[/tex]

The potential on the second plate

[tex]V_{2}=V-V_{1}[/tex]

[tex]V_{2}=51.5 -9.18[/tex]

[tex]V_{2}=42.32\ v[/tex]

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

Using formula of equivalent capacitance

[tex]C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}[/tex]

Put the value into the formula

[tex]C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}[/tex]

[tex]C=6.99\times10^{-7}\ F[/tex]

[tex]C=0.69\ \mu F[/tex]

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

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

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