Question

a charge of 17.7*10^-4C is distributed uniformly over a large sheet of area 200m^2 . Calculate the electric field intensity at a distance of 20cm from it in air.​

Answer 1

Answer:

Electric field intensity at 20 cm intensity is 5×10⁵ .

Explanation:

electric field intensity at a distance d from a large sheet is given as          

 $E = \frac{\sigma} {2\epsilon }$

where σ = surface charge density        

   ε = permittivity in free space

Given : q= 17.7* 10 ⁻⁴ C             area of sheet = 200m²             distance = 20 cm surface charge density is  

        σ$= \frac{q}{A}$  

  σ = $17.7 \times10^{-4} / 200 \times10 ^{2}$  

 σ = 8.85 × 10 ⁻⁸ Electric Field

 =  σ/ε   E = $\frac{8.85 *10^{-8} }{\epsilon\\}$

Hence E =5× 10⁵ NC⁻¹