Question

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the electric field at the origin. (Use ke, λ0, and x0 as necessary.)

Answer 1

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

Answer 2

Answer:

$E=-K_e\dfrac{\lambda_0 }{2x_0}N/C$

Explanation:

We know that electric field given as

$E=K_e\int_{x_1}^{x_2}\dfrac{\lambda }{x^2}dx$

Here given that

$\lambda =\dfrac{\lambda _0x_0}{x}$

So

$E=K_e\int_{x_1}^{x_2}\dfrac{\lambda_0x_0 }{x^3}dx$

$E=K_e\int_{x_0}^{\infty }\dfrac{\lambda_0x_0 }{x^3}dx$

Now by integrating we get

$E=-K_e\lambda_0\left [ \dfrac{1 }{2x^2} \right ]^{\infty }_x_0dx$

$E=-K_e\dfrac{\lambda_0 }{2x_0}dx$

So the electric field at the origin E

$E=-K_e\dfrac{\lambda_0 }{2x_0}N/C$