Chapter 17, Problem 64GP

To determine

Flow of charge in the battery.

Answer to Problem 64GP

The answer is $9.4\times {10}^{-8}C$ .

Explanation of Solution

Given data:

Capacitance, $C=2600pF=2600\times {10}^{-12}F$

Potential, $V=9.0V$

Formula used:

Capacitance for a parallel plate is $C={\text{ε}}_0\frac{A}{d}$ .

Where,

  $C$ is the capacitance of a capacitor.

  $d$ is the distance.

  $E$ is the electric field.

  ${\text{ε}}_{\text{0}}$ is the permittivity of free space.

Calculation:

The voltage across the capacitor stays constant since the capacitor remains connected to the battery.

Thus, the capacitance changes according to

  $C_{initial}={\text{ε}}_0\frac{A}{d}=2600\times {10}^{-12}F$

  $C_{final}=K\frac{{\text{ε}}_{0}A}{d}=K{C}_{initial}=5C_{initial}$

  $\begin{array}{c}{Q}_{final}-Q_{initial}=C_{final}V-C_{initial}V\\ =\left(5C_{initial}-C_{initial}\right)V\\ =4C_{initial}V\end{array}$

  $=4\left(2600\times {10}^{-12}F\right)\left(9.0V\right)$

  $=9.4\times {10}^{-8}C$

Conclusion:

The required value is $9.4\times {10}^{-8}C$ .