Video: Determining the Net Electric Field at a Point Near a Conducting Arc

A rod bent into the arc of a circle of radius 0.500 m subtends an angle 2πœƒ at the center 𝑃 of the circle as shown. Each πœƒ equals 30Β°. If the rod is charged uniformly with a total density 0.033 C/m, what is the magnitude of the electric field at 𝑃?

04:22

Video Transcript

A rod bent into the arc of a circle of radius 0.500 meters subtends an angle two πœƒ at the center 𝑃 of the circle as shown. Each πœƒ 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 𝑃?

Looking at our diagram, we see point 𝑃 at the center of the circle, where the rod that’s been bent into a circular arc subtends an angle of two times 30 degrees, or 60 degrees. We’ll call the magnitude of the electric field at point 𝑃 capital 𝐸. And we’ll 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; we’ll call it 𝑄. And at some other point, a distance we can say π‘Ÿ away, we want to know what the electric field created by that point charge 𝑄 is. That electric field magnitude is equal to one over four πœ‹πœ– naught, where πœ– naught is a constant, multiplied by the charge magnitude 𝑄 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, that’s fine for a point charge. But in our case, we have a distributed charge spread out over the arc of a circle. And that’s true. In the problem statement, we’re 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, we’ll see that we’re actually able to express the electric field at 𝑃 in terms of the electric field created by a point charge.

Let’s say that, for the moment, we’ll ignore this entire rod except for a tiny segment of the rod that carries a charge we’ll call 𝑑𝑄. This tiny amount of charge creates an electric field which can be experienced at point 𝑃. And based on our equation for the electric field created by a point charge, we could say that 𝑑𝐸 is equal to one over four πœ‹πœ– naught times 𝑑𝑄 over 𝑅 squared.

Looking back at our little segment of charge 𝑑𝑄, we know that 𝑑𝑄 will depend on the charge density πœ†. We can write that 𝑑𝑄 is equal to πœ† times 𝑑𝑠, 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 𝑠 is equal to π‘Ÿ, the circle’s radius, times πœƒ, the angular displacement of the arc, then that means we can write πœ†π‘‘π‘  as πœ† times capital 𝑅, the radius of our circle, times π‘‘πœƒ.

Making that substitution into our equation for 𝑑𝐸, we see that a factor of capital 𝑅, the radius of the circle, cancels. At this point, we now have a good expression for the electric field created at point 𝑃 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 we’ll integrate from angles of zero to two πœƒ radians and on the left-hand side we’ll simply integrate from zero to 𝐸, because the left-hand side will simplify to capital 𝐸. Knowing that πœƒ is given as 30 degrees and therefore two πœƒ is 60 degrees, we can say that the electric field at 𝑃 is equal to one over four πœ‹πœ– naught times πœ† over 𝑅 multiplied by that angle πœ‹ over three. That’s the equivalent in radians of 60 degrees.

Now that we’ve solved for the electric field at 𝑃 symbolically, it’s time to plug in and solve for it numerically. We’re told in the problem statement that the circle radius 𝑅 is 0.500 meters. And we can look up πœ– naught and find that it’s 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 𝐸 is 6.2 times 10 to the eighth newtons per coulomb. That’s the electric field created at point 𝑃 due to this curved rod.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.