Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance, as a result of their mutual attraction.

(a) Use Eq. 2.52 to express the work done by electrostatic forces, in terms of the fieldE, and the area of the plates, 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.)

Short Answer

Expert verified

(a) The work done by electrostatic forces in moving the plate is, W=12ε0εAE2.

(b) The energy lost by the fields in this process is W=12ε0AE2.

Step by step solution

01

Define the sketch for the condition.

Determine the sketch for the condition.


02

Determine work by electrostatic forces

(a)

Write the expression electrostatic pressure on either plate,

P=12ε0E2

Hence, the force on each plate is given by,

F=PA=12ε0E2A=12ε0AE2Towardstheotherplate

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

w=Fd=12ε0εAE2

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

w=Fd=12ε0εAE2

03

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=12ε0E2

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

W=12ε0E2(changethevolumeoccupiedbytheelectricalfieldbetweentheplates)=12ε0E2Ad-Ad-ε=12ε0AE2

Hence, the energy lost by the fields in this process isW=12ε0AE2

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