Question

A conductor of any shape, having area 40 cm² placed in air is uniformly charged with a charge 0* 2µC. Determine the electric intensity at a point just outside its surface. Also, find the mechanical force per unit area of the charged conductor.

[∈₀=8.85x10^-12 S.I. units]

Answer 1

Given: Q=0.2 μC=0.2*10^-6C

A=40cm²=40*10^-4m²

ε₀=8.85*10^-12SI units

The electric field intensity just outside the surface of a charged conductor of any shape is

`E=sigma/epsi_0=Q/(Aepsi_0)`

∴`E=(0.2*10^-6)/(40*10^-4*8.85*10^-12`

∴`E=5.65*10^6`N/C

Now, the mechanical force per unit area of a conductor is

`f=1/2epsi_0E^2=1/2*8.85*10^-12*(5.65*10^6)^2`

∴f=141.25N/m²