Chapter 25, Problem 1P

(a) When a battery is connected to the plates of a 3.00-μF capacitor, it stores a charge of 27.0 μC. What is the voltage of the battery? (b) If the same capacitor is connected to another batten and 36.0 μC of charge is stored on the capacitor, what is the voltage of the battery?

To determine

The voltage of the battery.

Answer to Problem 1P

The voltage of the battery is $9.00\text{V}$.

Explanation of Solution

Given information: Capacitance of the system is $3.00\text{μF}$, charge stored by the capacitor is $27.0\text{μC}$.

Formula to calculate the voltage of the battery.

$\begin{array}{c}Q=CV\\ V=\frac{Q}{C}\end{array}$ (1)

Here,

$V$ is the voltage of the battery.

$Q$ is the charge stored by the capacitor.

$C$ is the capacitance of the system.

Substitute $27.0\text{μC}$ for $Q$, $3.00\text{μF}$ for $C$ in equation (1) to find $V$,

$\begin{array}{c}V=\frac{27.0\text{μC}\times \frac{1\text{C}}{1000000\text{μC}}}{3.00\text{μF}\times \frac{1\text{F}}{1000000\text{μF}}}\\ =9.00\text{V}\end{array}$

Thus, the voltage of the battery is $9.00\text{V}$.

Conclusion:

Therefore, the voltage of the battery is $9.00\text{V}$.

To determine

The voltage of the another battery.

Answer to Problem 1P

The voltage of the another battery is $12.0\text{V}$.

Explanation of Solution

Given information: Capacitance of the system is $3.00\text{μF}$, charge stored by the capacitor is $36.0\text{μC}$.

Formula to calculate the voltage of the another battery.

$\begin{array}{c}Q=CV\\ V=\frac{Q}{C}\end{array}$ (2)

Here,

$V$ is the voltage of the another battery.

$Q$ is the charge stored by the capacitor.

$C$ is the capacitance of the system.

Substitute $36.0\text{μC}$ for $Q$, $3.00\text{μF}$ for $C$ in equation (2) to find $V$,

$\begin{array}{c}V=\frac{36.0\text{μC}\times \frac{1\text{C}}{1000000\text{μC}}}{3.00\text{μF}\times \frac{1\text{F}}{1000000\text{μF}}}\\ =12.0\text{V}\end{array}$

Thus, the voltage of the another battery is $12.0\text{V}$.

Conclusion:

Therefore, the voltage of the another battery is $12.0\text{V}$.