Question

A battery with an emf of 60 {\rm V} is connected to the two capacitors shown in the figure . Afterward, the charge on capacitor 2 is 540 \mu C.

What is the capacitance of capacitor 2? C₂ = ____μF

Express your answer using two significant figures.

(sorry the immage didnt work well but the 12 uF and C₂ are connected in the circuit it just wont line up for me)

Answer 1

Concepts and reason

The concepts required to solve the given question are the capacitance of the series combination of capacitors and the relation between the charge, voltage and the capacitance of a circuit.

First, calculate the charge on the capacitors which are connected in a series. Later use the relation between charge, voltage and capacitance of a circuit to find the equivalent capacitance of the capacitors. Finally, calculate the capacitance of the second capacitor using the expression for the equivalent capacitance of the capacitors that are connected in a series.

Fundamentals

Capacitance of a device may be defined as the “ability of a device to store charge”. Parallel plates that are connected to a battery can act as a capacitor. The battery provides the transportation of charge from one plate to another until the voltage produced across the plates is equal to the voltage of the battery.

One plate of the capacitor gets positively charged and the other plate gets negatively charged. These oppositely charged parallel plates produce a potential difference across it. It depends upon the capacitance and the charge stored in the capacitor.

Charge stored in the capacitor,

$Q = CV$

Here, V is the potential difference between the plates, Q is the charge stored in the plates and C is the capacitance of the parallel plate capacitor.

Capacitance of the series combination of capacitors is given by,

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Here, ${C_1}$ and ${C_2}$ are the capacitance of the capacitors 1 and 2 respectively, and ${C_{eq}}$ is the equivalent capacitance of the combination.

Equivalent charge of the capacitors connected in series:

The charge stored in the capacitors that are connected in a series,

$Q = {Q_1} = {Q_2}$

Here, $Q$ is the charge stored in the capacitors and ${Q_1}$ and ${Q_2}$ are the net charge stored in the capacitors 1 and 2 respectively.

Substitute $540{\rm{ \mu C}}$ for ${Q_2}$ .

$\begin{array}{c}\\Q = {Q_1}\\\\ = \left( {540{\rm{ \mu C}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ \mu C}}}}} \right)\\\\ = 540 \times {10^{ - 6}}{\rm{ C}}\\\end{array}$

The net charge in the circuit is given as,

$Q = V{C_{eq}}$

Here, V is the applied voltage to the circuit.

Rearrange the above equation for ${C_{eq}}$ .

${C_{eq}} = \frac{Q}{V}$

Substitute 60 V for V and $540 \times {10^{ - 6}}{\rm{C}}$ for $Q$ .

$\begin{array}{c}\\{C_{eq}} = \frac{{540 \times {{10}^{ - 6}}{\rm{ C}}}}{{60{\rm{ V}}}}\\\\ = 9 \times {10^{ - 6}}{\rm{ F}}\\\end{array}$

Capacitance of the unknown capacitor:

The equivalent capacitance of the series combination of capacitors is given as,

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Rearrange the above relation for ${C_2}$ ,

$\frac{1}{{{C_2}}} = \frac{1}{{{C_{eq}}}} - \frac{1}{{{C_1}}}$

Substitute $9 \times {10^{ - 6}}{\rm{F}}$ for ${C_{eq}}$ and $12\,{\rm{\mu F}}$ for ${C_1}$ .

$\begin{array}{c}\\\frac{1}{{{C_2}}} = \left( {\frac{1}{{9 \times {{10}^{ - 6}}{\rm{ F}}}}} \right) - \left( {\left( {\frac{1}{{12{\rm{ \mu F}}}}} \right)\left( {\frac{{1{\rm{ \mu F}}}}{{{{10}^{ - 6}}{\rm{ F}}}}} \right)} \right)\\\\ = 2.7778 \times {10^4}{\rm{ }}{{\rm{F}}^{ - 1}}\\\\{C_2} = \frac{1}{{\left( {2.7778 \times {{10}^4}{\rm{ }}{{\rm{F}}^{ - 1}}} \right)}}\\\\ = \left( {3.6 \times {{10}^{ - 5}}{\rm{ F}}} \right)\left( {\frac{{1{\rm{ \mu F}}}}{{{{10}^{ - 6}}{\rm{ F}}}}} \right)\\\\ = 36{\rm{ \mu F}}\\\end{array}$

Ans:

The capacitance of capacitor 2 is $36\,{\rm{\mu F}}$ .