Question

Suppose two thin flat plates measure 3.5m×3.5m and are separated by 5.5 mm . They are oppositely charged with ±18μC.

A) Estimate the total force exerted by one plate on the other (ignore edge effects).

B)How much work would be required to move the plates from 5.5

mm apart to 1.30 cm apart?

Answer 1

(A) Here suppose the followings -

Q: Charge on one plate

E: Electric field between the plates
∆V: Voltage drop across the plates
d: Distance separating the plates
εo: Permittivity of free space
A: Area of one plate

Now, neglecting edge effects, the force on one plate is

= Q·E

= Q[ ∆V / d ]

= Q[ (Q/C) / d ]

= Q² / Cd

= Q² / [ (εoA/d)d ]

= Q² / εoA

put in the values -

Q = 18 x 10^-6 C, A = 3.5 x 3.5 m^2

F = ( 18 × 10^-6 C )² / [ ( 8.85 × 10^-12 F/m )( 3.5 x 3.5 m² ) ]
= 3.0 N  

(B) The work separating this from 5.5 mm to 1.30 cm is

W = F∆d
= ( 3.0 N)( 0.013 m - 0.0055 m )
= 0.0225 J