A 12-nC charge is distributed uniformly along the y axis from y = 0 to y = 4 m. Which of the following integrals is correct for the x component of the electric field at x = 2 m on the x axis?
E) none of these

Question

Quizplus · Accepted Answer

To find the x component of the electric field at x = 2 m on the x-axis, we need to integrate the electric field contributions from each segment of the y-axis. We can start with Coulomb's law:

$\lvert\vec{E}vert = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

The y component of the electric field at a point on the x-axis is directed along the y-axis, and its magnitude is given by:

$\lvert E_y vert = \frac{1}{4\pi\epsilon_0}\frac{dq}{r^2}\sin	heta = \frac{1}{4\pi\epsilon_0}\frac{\lambda dy}{r^2}\sin	heta$

where $dq$ is a small element of charge along the y-axis, $\lambda$ is the charge density, $dy$ is its length, $r$ is the distance from the charge element to the point on the x-axis, and $	heta$ is the angle between the line joining the charge element and the point on the x-axis, and the y-axis. Since we are interested in the x component of the electric field, we need to find $\cos	heta$. This can be done by noting that $\cos	heta = \frac{x}{r}$ and $\sin	heta = \frac{y}{r}$, and that $r = \sqrt{x^2 + y^2}$. Thus, we have:

$\lvert E_x vert = \frac{1}{4\pi\epsilon_0}\frac{\lambda dy}{r^2}\cos	heta = \frac{1}{4\pi\epsilon_0}\frac{\lambda dy}{\sqrt{x^2+y^2}}\frac{x}{x^2 + y^2}$

where $\lambda = \frac{Q}{L}$ is the charge density, $Q$ is the total charge, and $L$ is the length of the charge distribution. To obtain the x component of the electric field at x = 2 m, we need to integrate this expression over the range of y values that the charge distribution covers, which is from 0 to 4 m. Therefore,

$E_x = \int_0^4 \frac{1}{4\pi\epsilon_0}\frac{\lambda}{\sqrt{x^2+y^2}}\frac{x}{x^2 + y^2}\, dy$

Substituting the values gives the answer as 6.91*10^7 N/C. Thus, the correct integral is B).

A 12-NC Charge Is Distributed Uniformly Along the Y Axis