Respuesta :
Explanation:
it is given that, the linear charge density of a charge, [tex]\lambda=\dfrac{\lambda_ox_o}{x}[/tex]
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 :
[tex]dE=\dfrac{k\ dq}{x^2}[/tex]..........(1)
The linear charge density is given by :
[tex]\lambda=\dfrac{dq}{dx}[/tex]
[tex]dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx[/tex]
Integrating equation (1) from x = x₀ to x = infinity
[tex]E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx[/tex]
[tex]E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}[/tex]
[tex]E=\dfrac{k\lambda_o}{2x_o}[/tex]
Hence, this is the required solution.
Answer:
[tex]E=-K_e\dfrac{\lambda_0 }{2x_0}N/C[/tex]
Explanation:
We know that electric field given as
[tex]E=K_e\int_{x_1}^{x_2}\dfrac{\lambda }{x^2}dx[/tex]
Here given that
[tex]\lambda =\dfrac{\lambda _0x_0}{x}[/tex]
So
[tex]E=K_e\int_{x_1}^{x_2}\dfrac{\lambda_0x_0 }{x^3}dx[/tex]
[tex]E=K_e\int_{x_0}^{\infty }\dfrac{\lambda_0x_0 }{x^3}dx[/tex]
Now by integrating we get
[tex]E=-K_e\lambda_0\left [ \dfrac{1 }{2x^2} \right ]^{\infty }_x_0dx[/tex]
[tex]E=-K_e\dfrac{\lambda_0 }{2x_0}dx[/tex]
So the electric field at the origin E
[tex]E=-K_e\dfrac{\lambda_0 }{2x_0}N/C[/tex]
