Question

A long cylindrical conductor of radius 2 cm carries a charge of 5 c/m and is kept in a medium of dielectric constant 10. Find the electric field intensity at a point 1 m from the axis of the cylinder.

Answer 1

The electric field intensity at a point 1 m from the axis of the cylinder is 8.99×10⁹ V-m

According to Gauss' law,

The total electric flux from a closed surface is equal to the charge enclosed divided by the permittivity.

So, ξ₀kE2Пdl = Charge enclosed

ξ₀ is the absolute permittivity = 8.85×10⁻¹² F/m

k is the dielectric constant = 10

E is the electric field

d is the distance of point from the axis of cylinder = 1m

Charge enclosed = Charge per unit length (λ) × Length (l)

= 5lC/m

So, E = (λl)/(ξ₀k2Пdl)

Substituting the values, we get

E = (5l)/(8.85×10⁻¹² × 10 ×2П ×1l)

= 5/(8.85×10⁻¹² × 10 ×2П ×1)

= 8.99×10⁹ V-m