Question

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E⃗ at the point P, a distance d above the midpoint of the wire.

Answer 1

Answer:

The electric field at point P is $2k\lambda d(\dfrac{L}{d^2\sqrt{L^2+d^2}})$

Explanation:

Given that,

Length = 2L units

Linear charge density = λ

We need to calculate the electric field at point P

Using formula of electric field

$E=2\int_{0}^{L}{k\dfrac{\lambda}{r^2}dx\sin\theta}$

Put the value into the formula

$E=2k\int_{0}^{L}{\dfrac{\lambda}{(k^2+d^2)}\times\dfrac{d}{\sqrt{x^2+d^2}}dx}$

$E=2k\lambda d\int_{0}^{L}{\dfrac{dx}{(x^2+d^2)^{\frac{3}{2}}}}$

$E=2k\lambda d(\dfrac{x}{d^2\sqrt{x^2+d^2}})_{0}^{L}$

$E=2k\lambda d(\dfrac{L}{d^2\sqrt{L^2+d^2}})$

Hence, The electric field at point P is $2k\lambda d(\dfrac{L}{d^2\sqrt{L^2+d^2}})$