Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

Answer to Problem 85P

  $U=\frac{Q^{2}d}{2{\in }_{0}a\left[\left(\kappa -1\right)x+a\right]}$

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor $=Q$

Distance between plated $=d$

Lengths of capacitor plates $=a$ and b

Formula used :

Energy stored in capacitor is:

  $U=\frac{1}{2}\frac{Q^2}{C}$

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

  $C_{eq}=C_1+C_2+\mathrm{...}$

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance $C_{eq}$ is,

  $U=\frac{1}{2}\frac{Q^2}{C_{eq}}$

The capacitances of the two capacitors are,

  $C_1=\frac{\kappa {\in }_{0}ax}{d}$

And

  $C_2=\frac{{\in }_{0}a\left(a-x\right)}{d}$

  $\begin{array}{l}{C}_{eq}=C_1+C_2\\ C_{eq}=\frac{\kappa {\in }_{0}ax}{d}+\frac{{\in }_{0}a\left(a-x\right)}{d}\\ C_{eq}=\frac{{\in }_{0}a}{d}\left(\kappa x+a-x\right)\\ C_{eq}=\frac{{\in }_{0}a}{d}\left[\left(\kappa -1\right)x+a\right]\end{array}$

  $U=\frac{Q^{2}d}{2{\in }_{0}a\left[\left(\kappa -1\right)x+a\right]}$

Conclusion:

The energy stored in the capacitor is, $U=\frac{Q^{2}d}{2{\in }_{0}a\left[\left(\kappa -1\right)x+a\right]}$

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

Answer to Problem 85P

  $F=\frac{\left(\kappa -1\right)Q^{2}d}{2{\in }_{0}a{\left[\left(\kappa -1\right)x+a\right]}^2}$

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor $=Q$

Distance between plated $=d$

Lengths of capacitor plates $=a\text{and}b$

Energy stored in the capacitor, $U=\frac{Q^{2}d}{2{\in }_{0}a\left[\left(\kappa -1\right)x+a\right]}$

Formula used:

Electric force:

  $F=-\frac{dU}{dx}$

U is the energy stored.

Calculation:

  $\begin{array}{l}F=-\frac{dU}{dx}\\ F=-\frac{d}{dx}\left[\frac{1}{2}\frac{Q^2}{{\in }_{0}a\left[\left(\kappa -1\right)x+a\right]}\right]\\ F=\frac{Q^{2}d}{2{\in }_{0}a}\frac{d}{dx}\left\{{\left[\left(\kappa -1\right)x+a\right]}^{-1}\right\}\\ F=\frac{\left(\kappa -1\right)Q^{2}d}{2{\in }_{0}a{\left[\left(\kappa -1\right)x+a\right]}^2}\end{array}$

Conclusion:

The force by examining how the stored energy varies with x is,

  $F=\frac{\left(\kappa -1\right)Q^{2}d}{2{\in }_{0}a{\left[\left(\kappa -1\right)x+a\right]}^2}$

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

Answer to Problem 85P

  $F=\frac{\left(\kappa -1\right)a{\in }_0{V}^2}{2d}$

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor $=Q$

Distance between plated $=d$

Lengths of capacitor plates $=a\&b$

  $F=\frac{\left(\kappa -1\right)Q^{2}d}{2{\in }_{0}a{\left[\left(\kappa -1\right)x+a\right]}^2}$

  $C_{eq}=\frac{{\in }_{0}a}{d}\left[\left(\kappa -1\right)x+a\right]$

Formula Used:

The capacitance can be obtained by:

  $C=\frac{{\in }_{0}A}{d}$

Where, ${\in }_0$ is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, $Q=CV$

Where, C is the capacitance and V is the voltage.

Calculation:

  $F=\frac{\left(\kappa -1\right)Q^{2}d}{2{\in }_{0}a{\left[\left(\kappa -1\right)x+a\right]}^2}$

Multiply and divide by $\frac{a{\in }_o}{d^2}$ .

  $\begin{array}{l}F=\frac{\left(\kappa -1\right)Q^2\left(\frac{a{\in }_0}{d}\right)}{2{\left(\frac{a{\in }_0}{d}\right)}^2{\left[\left(\kappa -1\right)x+a\right]}^2}\\ F=\frac{\left(\kappa -1\right)Q^2\left(\frac{a{\in }_0}{d}\right)}{2C_{eq}^2}\\ Q=CV\\ \because C=\frac{{\in }_{0}A}{d}\\ Q=\left(\frac{{\in }_{0}A}{d}\right)V\\ F=\frac{\left(\kappa -1\right)a{\in }_0{V}^2}{2d}\end{array}$

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

  $F=\frac{\left(\kappa -1\right)a{\in }_0{V}^2}{2d}$

(d)

To determine

ToFind: The position from where the force originates.

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.