Question

Given a 2.75 µF capacitor, a 6.00 µF capacitor, and a 9.00 V battery, find the charge on each capacitor if you connect them in the following ways.
(a) in series across the battery
µC

(b) in parallel across the battery
2.75 µF capacitor µC
6.00 µF capacitor µC

Answer 1

Concepts and reason

The concepts that are to be used to solve the given problem are the combination of capacitors in parallel, and the charge that the capacitors store in a series and parallel combination.

First, calculate the equivalent capacitance of the capacitors by using the series combination of two capacitors. Next, calculate the charge on each capacitor in series combination by using the relation between charge and capacitance. Finally, calculate the charge on each capacitor in parallel combination across the battery.

Fundamentals

The equivalent capacitance in a series combination is,

$\begin{array}{l}\\\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}\\\\{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}\\\end{array}$

Here, ${C_{eq}}$ is the equivalent capacitance, ${C_1}$ is the capacitance of the capacitor 1, and ${C_2}$ is the capacitance of the capacitor 2.

The amount of charge acquired by a given capacitor is,

$Q = CV$

Here, $Q$ is the amount of charge, $V$ is the potential difference, and $C$ is the capacitance.

(a)

The equivalent capacitance for series combination of two capacitors,

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

Substitute $2.75 \times {10^{ - 6}}{\rm{ F}}$ for${C_1}$ and $6.00 \times {10^{ - 6}}{\rm{ F}}$ for ${C_2}$ in the equation${C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}.$

$\begin{array}{c}\\{C_{eq}} = \frac{{\left( {2.75 \times {{10}^{ - 6}}{\rm{ F}}} \right)\left( {6.00 \times {{10}^{ - 6}}{\rm{ F}}} \right)}}{{\left( {2.75 \times {{10}^{ - 6}}{\rm{ F}}} \right) + \left( {6.00 \times {{10}^{ - 6}}{\rm{ F}}} \right)}}\\\\ = 1.8857 \times {10^{ - 6}}{\rm{ F}}\\\end{array}$

The capacitors can store the same charge in a series combination.

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

Here, ${C_{eq}}$ is the equivalent capacitance and $\Delta V$ is the potential difference.

Substitute $1.8857 \times {10^{ - 6}}{\rm{ F}}$ for ${C_{eq}}$ and 9.00 V for $\Delta V$ in the equation, $Q = {C_{eq}}\Delta V$.

$\begin{array}{c}\\Q = \left( {1.8857 \times {{10}^{ - 6}}{\rm{ F}}} \right)\left( {9.00{\rm{ V}}} \right)\\\\ = 16.9713 \times {10^{ - 6}}{\rm{ C}}\\\\ = 16.9713{\rm{ \mu C}}\\\end{array}$

(b)

The capacitors have same potential difference in the parallel combination.

$\begin{array}{l}\\{Q_1} = {C_1}\Delta V\\\\{Q_2} = {C_2}\Delta V\\\end{array}$

Here, ${Q_1}$ and ${Q_2}$ are the charges for the capacitors 1 and 2. ${C_1}$ and ${C_1}$ are the capacitance of the capacitors 1 and 2, and $\Delta V$ is the equal potential difference for two capacitors.

The charge stored on the capacitor 1 in a parallel combination is${Q_1} = {C_1}\Delta V$.

Substitute $2.75 \times {10^{ - 6}}{\rm{ F}}$ for C₁ and 9.00 V for $\Delta V$ in${Q_1} = {C_1}\Delta V$.

$\begin{array}{c}\\{Q_1} = \left( {2.75 \times {{10}^{ - 6}}{\rm{ F}}} \right)\left( {9.00{\rm{ V}}} \right)\\\\ = 24.75 \times {10^{ - 6}}{\rm{ C}}\\\\ = 24.75{\rm{ }}\mu {\rm{C}}\\\end{array}$

The charge stored on the capacitor 2 in a parallel combination is${Q_2} = {C_2}\Delta V$.

Substitute $6.00 \times {10^{ - 6}}{\rm{ F}}$ for C₂ and 9.00 V for $\Delta V$ in${Q_2} = {C_2}\Delta V$.

$\begin{array}{c}\\{Q_2} = \left( {6.00 \times {{10}^{ - 6}}{\rm{ F}}} \right)\left( {9.00{\rm{ V}}} \right)\\\\ = 54 \times {10^{ - 6}}{\rm{ C}}\\\\ = 54{\rm{ }}\mu {\rm{C}}\\\end{array}$

Ans: Part a

The charge on each capacitor those are connected in a series across the battery

µC is$16.9713{\rm{ \mu C}}.$

Part b

The charge stored on the capacitor 1 in parallel combination is $24.75{\rm{ }}\mu {\rm{C}}$ and on capacitor 2 is$54{\rm{ }}\mu {\rm{C}}.$