A conductor of any shape, having area 40 cm^{2} 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]

Advertisement Remove all ads

#### Solution

Given: Q=0.2 μC=0.2*10^{-6}C

A=40cm^{2}=40*10^{-4}m^{2}

ε_{0}=8.85*10^{-12}SI 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/m^{2}

Concept: Mechanical Force on Unit Area of a Charged Conductor

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads