Question

A wire with uniform charge distribution along its length with value of linear charge density 50 cm^ -1. What will be value of electric field strength E on 20 cm distance from it.​

Answer 1

Given:

A wire with uniform charge distribution along its length with value of linear charge density 50 C/m.

To find:

Value of Electrostatic Field Intensity at 20 cm distance from it ?

Calculation:

The general expression for electric field intensity from an infinite wire at a specified distance is given as :

$\boxed{ \bf \: E = \dfrac{2k \lambda}{r} }$

• k refers to Coulomb's Constant , $\lambda$ refers to linear charge density

• r refers to distance of point from wire.

Putting the available values in SI unit :

$\sf \implies\: E = \dfrac{2 \times (9 \times {10}^{9} ) \times 50 }{ (\frac{20}{100} )}$

$\sf \implies\: E = \dfrac{2 \times (9 \times {10}^{9} ) \times 50 }{0.2}$

$\sf \implies\: E = \dfrac{2 \times (9 \times {10}^{9} ) \times 500 }{2}$

$\sf \implies\: E = 9 \times {10}^{9} \times 500$

$\sf \implies\: E = 45 \times {10}^{11} \: N/C$

So, the field intensity is 45 × 10¹¹ N/C.