Answer :
Answer:
Please refer to the figure.
We will use the following formula for the electric field:
[tex]|E| = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}[/tex].
Since this is not a point source, but a uniformly distributed charge, we will use the method of integration.
We will first choose an infinitesimal (very very small) point on the distribution.
The infinitesimal length of the point is denoted as [tex]dy[/tex], and the charge of this point is denoted as [tex]dQ[/tex].
Since the whole distribution is uniform, we can write the charge density as follows: [tex]\lambda = \frac{Q}{Length} = \frac{Q}{4}[/tex].
The small fraction (dQ) of this distribution should have the same charge density (E.g.: a bucket of water and a glass of the same water have the same mass density). [tex]\lambda = \frac{dQ}{dy}[/tex].
This yields
[tex]dQ = \frac{Q}{4} dy[/tex]
Now we can ignore the rest of the distribution and focus only the point charge (dQ) and plug this into the electric field equation:
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(y^2 + 4)}[/tex]
where [tex](y^2 + 4)[/tex] is the distance from the charge to the x = 2 on the x-axis.
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{4}\frac{dy}{(y^2+4)}[/tex]
x-component of dE is [tex]dE_{x} = dE*\cos(\alpha) = dE*\frac{2}{\sqrt(y^2+4))}[/tex]
[tex]dE_{x} = \frac{1}{4\pi\epsilon_0}\frac{Q}{4}\frac{2dy}{(y^2+4)^{3/2} }[/tex]
This is the x component of the point charge (dQ).
Now we will integrate this quantity over the whole distribution to obtain the x component of the electric field at x = 2m on the x axis:
[tex]E_{x} = \frac{1}{4\pi\epsilon_0}\frac{Q}{2} \int\limits^4_0 {\frac{1}{(y^2+4)^{3/2}}} \, dy[/tex].
Since [tex]Q = 12nC[/tex];
The answer should look like
[tex]E_{x} = \frac{1}{4\pi\epsilon_0} \int\limits^4_0 {\frac{6}{(y^2+4)^{3/2}}} \, dy[/tex]
Explanation:
The only equation we know regarding this question is the electric field equation, but that equation is only valid for point charges. However, this question proposes a uniformly distributed charge, so all we have to do is to adapt the given system to what we know: Point charge equation of electric field.
The way to do this is to use this useful information: A line is consist of the sum of infinitely many points. We can use integral to sum over all points, so we just have to choose a representative point, and apply the electric field equation to that point.
