Chapter 19, Problem 45P

(a)

To determine

The potential difference across each capacitor.

Answer to Problem 45P

The potential difference across first capacitor is 5.4 V and the potential difference across second capacitor is 3.6 V.

Explanation of Solution

Given info:

The capacitance of the first capacitor is,

  $C_1=0.40\text{μF}$ .

The capacitance of the second capacitor is, $C_2=0.60\text{μF}$ .

The voltage of the battery is, $V=9\text{V}$ .

Formula Used:

The expression to calculate the potential difference across first capacitor is,

  $V_1=\left(\frac{C_2}{C_1+C_2}\right)V$

The expression to calculate the potential difference across second capacitor is,

  $V_2=\left(\frac{C_1}{C_1+C_2}\right)V$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{V}_1=\left(\frac{0.60\text{μF}}{0.40\text{μF}+0.60\text{μF}}\right)\left(9\text{V}\right)\\ =5.4\text{V}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{V}_2=\left(\frac{0.40\text{μF}}{0.40\text{μF}+0.60\text{μF}}\right)\left(9\text{V}\right)\\ =3.6\text{V}\end{array}$

Conclusion:

Therefore, the potential difference across first capacitor is 5.4 V and the potential difference across second capacitor is 3.6 V.

(b)

To determine

The charge on each capacitor.

Answer to Problem 45P

The charge on the first capacitor is $3.6\text{μC}$ and the charge on the second capacitor is $5.4\text{μC}$ .

Explanation of Solution

Formula Used:

The expression to calculate the charge on the first capacitor is,

  $Q_1=C_{1}V$

The expression to calculate the charge on the second capacitor is,

  $Q_2=C_{2}V$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{Q}_1=\left(0.40\text{μF}\right)\left(9\text{V}\right)\\ =3.6\text{μC}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{Q}_2=\left(0.60\text{μF}\right)\left(9\text{V}\right)\\ =5.4\text{μC}\end{array}$

Conclusion:

Therefore, the charge on the first capacitor is

  $3.6\text{μC}$ and the charge on the second capacitor is

  $5.4\text{μC}$ .

(c)

To determine

The potential difference across each capacitor and charge on the each capacitor for parallel combinations of the capacitors.

Answer to Problem 45P

The charge on the first capacitor is $3.6\text{μC}$ and the charge on the second capacitor is $5.4\text{μC}$ and the potential difference across each capacitor is 9 V.

Explanation of Solution

Formula Used:

Both the capacitors are in the parallel combination, so potential difference across each capacitor in parallel combination remains same. The potential difference across each capacitor is 9 V.

The expression to calculate the charge on the first capacitor is,

  $Q_1=C_{1}V$

The expression to calculate the charge on the second capacitor is,

  $Q_2=C_{2}V$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{Q}_1=\left(0.40\text{μF}\right)\left(9\text{V}\right)\\ =3.6\text{μC}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{Q}_2=\left(0.60\text{μF}\right)\left(9\text{V}\right)\\ =5.4\text{μC}\end{array}$

Conclusion:

Therefore, the charge on the first capacitor is $3.6\text{μC}$ and the charge on the second capacitor is $5.4\text{μC}$ and the potential difference across each capacitor is 9 V.