Answer:
Explanation:
From the image attached below:
It is required to determine the electric field because of the line charge distribution of positive charge distributed uniformly from x = 0 to x = a at point x from O;
where x > a
Assume we chose an element dy at a distance from the point.
Then, the change on [tex]dy = \dfrac{Q}{a} \times dy[/tex]
The electric field due to this dy length is [tex]\dfrac{kdq}{y^2 }= \dfrac{kQ dy }{ay^2}[/tex]
Thus, the total electric field [tex]= \dfrac{kQ}{a}\int \limits ^{x}_{x-a} \dfrac{dy}{y^2}[/tex]
[tex]= \dfrac{kQ}{a} \Big [ \dfrac{1}{y}\Big ]^{x-a}_{x}[/tex]
[tex]=\dfrac{kQ}{a}\Big [\dfrac{1}{x-a}-\dfrac{1}{x}\Big ][/tex]
[tex]E=\dfrac{kQ}{a}\Big (\dfrac{x-(x-a)}{(x-a)x}\Big )[/tex]
[tex]E=\dfrac{kQa}{a(x-a)x}[/tex]
Hence, the electric field [tex]E=\dfrac{kQ}{(x-a)x} \ \ \ where; x>a[/tex]