Question

An infinite lineof charge with charge density λ1 = -0.2 μC/cm is aligned with the y-axis as shown.

1)What is Ex(P), the value of the x-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 7.7cm?____N/C
2)What is Ey(P), the value of the y-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 7.7cm?_____N/C

Answer 1

Just apply Gauss Law here, i.e Net Flux = close int(E.dS) = Net charge enclosed / permitivity
So we can apply this law or use the integration part to find flux.
By using the law, net charge enclosed is ?1 * h. Divide this by permitivity. There you go!

Another method is from observing elec. field.
In the integral E.dS , both E and dS are vectors, and we integrate their scalar product.
So the E due to wire is perpendicular to the curved face of cylinder, and so is the dS vector(Area normal) of the curved surface. So their scalar product is simply arithematic product.
Also, value of E at every point on curved surface is same.
So flux is simply (E)(Curved surface area)
Now E in this case is ?1/2*pi*a*permitivity and area is 2*pi*a*h
You again get flux as ?1*h/permitivity.