Chapter 17, Problem 75GP

(a)

To determine

The charge on the capacitor.

Answer to Problem 75GP

The charge on the capacitor is $4.24\times {10}^{-11}\text{C}$ .

Explanation of Solution

Given info:

Plate area is $A=2{\text{cm}}^{\text{2}}$ .

Air gap separation is $d=0.5\text{mm}$ .

Voltage across the plate is $\Delta V=12\text{V}$ .

Formula used:

The charge on the capacitor is,

  $Q=\frac{{\epsilon }_{0}A}{d}\times \Delta V$

Here, ${\epsilon }_0$ is the absolute permittivity.

Calculation:

Substituting the given values,

  $\begin{array}{c}Q=\frac{2\times {10}^{-4}}{\left(4\pi \times 9\times {10}^9\right)\left(0.5\times {10}^{-3}\right)}\times \left(12\right)\text{C}\\ \approx 4.24\times {10}^{-11}\text{C}\end{array}$

Conclusion:

Thus, the charge on the capacitor is $4.24\times {10}^{-11}\text{C}$ .

(b)

To determine

The charge on the capacitor if plates are now pulled to a separation of $0.75\text{mm}$ .

Answer to Problem 75GP

The charge on the capacitor is $4.24\times {10}^{-11}\text{C}$ .

Explanation of Solution

Given info:

Plate area is $A=2{\text{cm}}^{\text{2}}$ .

Air gap separation is $d=0.75\text{mm}$ .

Voltage across the plate is $\Delta V=12\text{V}$ .

Formula used:

The capacitance is,

  $C=\frac{{\epsilon }_{0}A}{d}$

Here, ${\epsilon }_0$ is the absolute permittivity.

Calculation:

Substituting the given values

  $\begin{array}{c}C=\frac{2\times {10}^{-4}}{\left(4\pi \times 9\times {10}^9\right)\left(0.75\times {10}^{-3}\right)}\\ \approx 2.36\times {10}^{-12}\text{F}\end{array}$

But the plate separation does not affect the charge stored as it is conserved.

Conclusion:

Thus, the charge on the capacitor is $4.24\times {10}^{-11}\text{C}$ .

(c)

To determine

The potential difference across the plate.

Answer to Problem 75GP

The potential difference across the plate is $17.98\text{V}$ .

Explanation of Solution

Given info:

Charge is $Q=4.24\times {10}^{-11}\text{C}$

The capacitance is $C=2.36\times {10}^{-12}\text{F}$

Formula used:

The potential difference across the plate is,

  $\Delta V\text{'}=\frac{Q}{C}$

Calculation:

Substituting the given values

  $\begin{array}{c}\Delta V\text{'}=\frac{4.24\times {10}^{-11}\text{C}}{2.36\times {10}^{-12}\text{F}}\\ =17.98\text{V}\end{array}$

Conclusion:

Thus, the potential difference across the plate is $17.98\text{V}$ .

(d)

To determine

The work required to pull the plates to their new separation.

Answer to Problem 75GP

The work done in separating the plates is $1.27\times {10}^{-10}\text{J}$ .

Explanation of Solution

Given info:

Charge is $Q=4.24\times {10}^{-11}\text{C}$

Voltage across the plate is $\Delta V=12\text{V}$ .

The potential difference across the plate is $\Delta V\text{'}=17.98\text{V}$ .

Formula used:

The work done in separating the plates is,

  $W=\frac{1}{2}Q\left(\Delta V\text{'}-\Delta V\right)$

Calculation:

Substituting the given values,

  $\begin{array}{c}W=\frac{1}{2}\left(4.24\times {10}^{-11}\text{C}\right)\left(17.98\text{V}-12\text{V}\right)\\ =1.27\times {10}^{-10}\text{J}\end{array}$

Conclusion:

Thus, the work done in separating the plates is $1.27\times {10}^{-10}\text{J}$ .