Q41P

Question

Two large metal plates (each of area A) are held a small distance d

a part. Suppose we put a chargeQon each plate; what is the electrostatic pressure on the plates?

Step-by-Step Solution

Verified
Answer

The electrostatic pressure on each plate of the parallel plate capacitor is  P=Q22ε0A2

1Step 1: Determine the electric field.

The electric field on either side of a charge-density conducting plate is as follows:

E=σ2ε0


Here, ε0 is the permittivity for free space, E is the electric field and σ charge density.


2Step 2: Determine magnitude of electric force

The magnitude of the electric force Facting on a charge in an electric field is expressed as,

F=EQ


Here,Qis the amount of the charge.

Substitute σ2ε0for E in above equation.


F=σ2ε0Q


Write the expression for surface charge density on each plate of the capacitor is,

σ=QA


Here, Ais the area of the surface.


SubstituteQA for σ in the equationF=σ2ε0Q


F=QA2ε0Q


=Q22ε0A

3Step 3: Determine the electrostatic pressure on the plates

The following formula can be used to calculate the pressure P acting on a surface due to a force F:

P=FA

Substitute Q22ε0A  for F


Determine the electrostatic pressure on the parallel plate capacitor's individual plates.

P=Q22ε0AA=Q22ε0A

Hence, the electrostatic pressure on each plate of the parallel plate capacitor is


P=Q22ε0A2