Positive charge Q is distributed uniformly along the x-axis...

Positive charge $Q$ is distributed uniformly along the $x$-axis from $x = 0$ to $x =$ a. A positive point charge $q$ is located on the positive $x$-axis at $x = a + \space r$, a distance $r$ to the right of the end of $Q \space (\textbf{Fig. P21.79}$). (a) Calculate the $x$- and $y$-components of the electric field produced by the charge distribution $Q$ at points on the positive $x$-axis where $x \space >$ a. (b) Calculate the force (magnitude and direction) that the charge distribution $Q$ exerts on $q$. (c) Show that if $r \space \gg a$, the magnitude of the force in part (b) is approximately $Qq/4\pi\epsilon_0 r^2$. Explain why this result is obtained.

Transcript

00:01 So here we have a rod from x equals to 0 to x equals to a.

00:07 So let's draw the axis first.

00:14 So let's say this is our x -axis and this is, excuse me, our y -axis.

00:23 And this is the negative x and negative y.

00:25 And we have a rod from x equals to 0 to x equals to a.

00:37 And the charge in the rod is uniformly distributed over the whole length of the rod.

00:55 And the information that we are given is the length, which is a, and the charge on the rod, which is denoted by capital q.

01:15 Now in the first part, we have to find the x and y components of the electric field.

01:21 So basically we have to find the electric field of this rod.

01:32 So let's do that.

01:33 Now, at any point on the x -axis, the vertical component of the electric field is 0.

01:46 And one more thing, we have to find the x and y components of the electric field produced by the charge distribution at points on the positive x -axis where x is greater than a.

01:58 So we have to find the electric field at any point outside this distance.

02:03 So let's say at this point here.

02:05 So let's call this point to be p.

02:09 And let's assume that p is placed at a distance x from the origin.

02:20 And we have to find the electric field at this point p.

02:23 Now again, as i said, at any point on the x axis, the vertical component of this electric field exerted by this charge distribution is 0.

02:37 Now let's find the linear charge distribution lambda of this rod.

02:42 So lambda will be equal to tq over dl.

02:48 So lambda is actually charge over since charge is uniformly distributed, so we can directly divide the total charge over the total length.

03:01 So this gives lambda to be q over a.

03:05 Now since charge is uniformly distributed throughout the rod, we can't just take directly the middle point or the whole point and find the electric field.

03:15 We have to cut the rod into small sections of small charges and find the electric field of that particular small charge at this point b.

03:26 So let's say we take this section and let's call say that this section is dl.

03:34 So this means this is at a distance l from the origin, which means that the rest of the distance, this distance is a minus l.

03:51 So the distance of dl from p is equal to minus l.

04:21 So this is the distance of dl from p.

04:26 Now the horizontal component, horizontal differential component, that is the horizontal component of electric field due to this segment or this small charge residing in this small segment of the rod will be equal to...

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