Q44P - Introduction to Electrodynamics

Question

 Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance${}^{\mathbf{\in }}$, as a result of their mutual attraction.

(a) Use Eq. 2.52 to express the work done by electrostatic forces, in terms of the field ${}^{\mathbf{E}}$, and the area of the plates, ${}^{\mathbf{A}}$.

(b) Use Eq. 2.46 to express the energy lost by the field in this process.

(This problem is supposed to be easy, but it contains the embryo of an alternative derivation of Eq. 2.52, using conservation of energy.)

Step-by-Step Solution

Answer

(a)  The work done by electrostatic forces in moving the plate is, $W=\frac{1}{2}{\epsilon }_0\epsilon A{E}^2$.

(b)  The energy lost by the fields in this process is $W=\frac{1}{2}{\epsilon }_{0}A{E}^2$.

Step 1: Define the sketch for the condition.

Determine the sketch for the condition. 

Step 2: Determine work by electrostatic forces

(a)

Write the expression electrostatic pressure on either plate,

$P=\frac{1}{2}{\epsilon }_0{E}^2$

 Hence, the force on each plate is given by,

$$\begin{gathered} F=PA \\ =\left(\frac{1}{2}{\epsilon }_0{E}^2\right)A \\ =\frac{1}{2}{\epsilon }_{0}A{E}^2 \\ \left(\mathrm{Towards} \mathrm{the} \mathrm{other} \mathrm{plate}\right) \end{gathered}$$

Write the expression for work done by electrostatic forces in moving the plate is,

$$\begin{gathered} w=Fd \\ =\frac{1}{2}{\epsilon }_0\epsilon A{E}^2 \end{gathered}$$

Thus, work done by electrostatic forces in moving the plate is, .

$$\begin{gathered} w=Fd \\ =\frac{1}{2}{\epsilon }_0\epsilon A{E}^2 \end{gathered}$$

Step 3: Determine the energy lost by the field in this process

(b)

Write the expression for Energy stored per unit volume in electrostatic field is given by,

$E=\frac{1}{2}{\epsilon }_0{E}^2$

Thus, energy lost by the fields in this process is given by,

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{E}}^{\mathbf{2}} (change the volume occupied by the electrical field between the plates) \\ =\frac{1}{2{\epsilon }_0}{E}^2\left[Ad-A\left(d-\epsilon \right)\right] \\ =\frac{1}{2}{\epsilon }_{0}A{E}^2 \end{gathered}$$

 Hence, the energy lost by the fields in this process is $W=\frac{1}{2}{\epsilon }_{0}A{E}^2$