### Video Transcript

A rod bent into the arc of a circle of radius 0.500 meters subtends an angle two π at the center π of the circle as shown. Each π equals 30 degrees. If the rod is charged uniformly with a total density 0.033 coulombs per meter, what is the magnitude of the electric field at π?

Looking at our diagram, we see point π at the center of the circle, where the rod thatβs been bent into a circular arc subtends an angle of two times 30 degrees, or 60 degrees. Weβll call the magnitude of the electric field at point π capital πΈ. And weβll start to solve for it by recalling the electric field of a point charge.

Imagine that we had at some point in space a point charge; weβll call it π. And at some other point, a distance we can say π away, we want to know what the electric field created by that point charge π is. That electric field magnitude is equal to one over four ππ naught, where π naught is a constant, multiplied by the charge magnitude π over the square of the distance between that charge and the point where we wanna solve for the field.

Looking at this relationship, we may think, well, thatβs fine for a point charge. But in our case, we have a distributed charge spread out over the arc of a circle. And thatβs true. In the problem statement, weβre given a linear charge density of this rod that was bent into a circular arc. But if we think on an infinitesimally small scale, as we consider the charge on this rod, say a very small portion at a time, weβll see that weβre actually able to express the electric field at π in terms of the electric field created by a point charge.

Letβs say that, for the moment, weβll ignore this entire rod except for a tiny segment of the rod that carries a charge weβll call ππ. This tiny amount of charge creates an electric field which can be experienced at point π. And based on our equation for the electric field created by a point charge, we could say that ππΈ is equal to one over four ππ naught times ππ over π
squared.

Looking back at our little segment of charge ππ, we know that ππ will depend on the charge density π. We can write that ππ is equal to π times ππ , a differential arc length segment of our curved rod. And if we recall that, for a circle, when we consider an arc length of that circle, that arc length π is equal to π, the circleβs radius, times π, the angular displacement of the arc, then that means we can write πππ as π times capital π
, the radius of our circle, times ππ.

Making that substitution into our equation for ππΈ, we see that a factor of capital π
, the radius of the circle, cancels. At this point, we now have a good expression for the electric field created at point π due to a very small segment of our curved rod. But of course we wanna calculate the electric field created by the entire length of the rod.

To do this, we can integrate our expression, where on the right-hand side of our expression weβll integrate from angles of zero to two π radians and on the left-hand side weβll simply integrate from zero to πΈ, because the left-hand side will simplify to capital πΈ. Knowing that π is given as 30 degrees and therefore two π is 60 degrees, we can say that the electric field at π is equal to one over four ππ naught times π over π
multiplied by that angle π over three. Thatβs the equivalent in radians of 60 degrees.

Now that weβve solved for the electric field at π symbolically, itβs time to plug in and solve for it numerically. Weβre told in the problem statement that the circle radius π
is 0.500 meters. And we can look up π naught and find that itβs 8.85 times 10 to the negative 12 farads per meter. Plugging in all these values and entering them on our calculator, to two significant figures, we find that πΈ is 6.2 times 10 to the eighth newtons per coulomb. Thatβs the electric field created at point π due to this curved rod.