Charge Q is uniformly distributed along a thin, flexible rod o...

Question

Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in FIGURE P23.48. Find an expression for the electric field Ē at the center of the semicircle. Hint: A small piece of arc length Δs spans a small angle Δθ=Δs/R , where R is the radius.

A semicircular rod with uniform charge distribution, showing the center point and arc length labeled.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with electric fields. So in this problem, we have a charge Q that is used formally distributed across the thin circumference of a half of a ring with radius R. As shown in the figure below, we need to determine the electric field in terms of Q and R at the center of the ring. In the figure below, we're given a coordinate system with the X and Y axis which the origin of that coordinate system corresponds to the center of a semicircle. And the semi circle is located below the X axis with each end of the semicircle touching the ax axis. We're given four possible choices as our answers. Choice A is 1/4 pi epsilon knot multiplied by Q divided by pi R squared multiplied by I hat. For choice B we have 1/4 pi epsilon knot multiplied by Q divided by pi R squared multiplied by J hat. For choice C we have 1/4 pi epsilon not multiplied by two Q divided by pi R squared multiplied by I hat. And for choice D, we have 1/4 pi epsilon N multiplied by two Q divided by pi R squared multiplied by J hat. Now, since we need to calculate the electric field and we have a continuous charge distribution, we need to recall our formula for the uh electric field for continuous charge distribution. So in that case, we have E is equal to the integral of one divided by four pi epsilon knot multiplied by DQ divided by R squared. And that's multiplied by R hat. Now, in this formula, um DQ represents a small charge element and R represents the distance from that charge element to the point of interest where we're trying to calculate the electric field. So if we go over to our diagram and we can look at what RDQ represents, it represents the charge on a small sliver of the arc. And if I look at the electric field generated by that DQ, if I look at this point on the right side of the semicircle, that's gonna cause an electric field to be pointing up into the left, I can find a similar point on the left side of the arc, we'll have another DQ and that will generate an electric field that's pointing up into the right. Now with these two electric fields, they have the same X component except one is going to the right and one's going to the left, the X components of the electric fields will cancel out. And I can do this for every point pair um around this arc. So that means that the only electric field um component that's going to contribute to my overall electric field is going to be the Y component. And so in order to calculate the Y component, I need to define this angle theta between the X axis and the electric field. So if I do that, I can rewrite my electric field formula as E is equal to the integral of one divided by or pi A not multiplied by DQ divided by R squared. And the place of R hat, I will have line of the multiplied by J hat because I know that we're only gonna have a Y component and that's in the J hat direction. I need to figure out what my DQ actually represents in this diagram. And we've already drawn what DQ looks like on the figure. And so DQ, there is just gonna be equal to the charge per unit length which will label as lambda. And that's gonna get multiplied by some small length. In this case is going to be a small arc length. We'll call it DS. We'll have DQ is equal to while the charge per unit length is going to be the total charge Q divided by the total arc length there, which I just have semi circle. So the circumference of this semi circle is just gonna be PI R and for the DS, I'm just gonna have a small arc length that's just gonna be RD theta, the art will cancel. And so my DQ, it's gonna be equal to you divided by pi D theta. And so I can plug that into my expression or the electric field, they have the electric field equal to the integral of one divided by four pi ute not we have with the DQ will have the Q divided by pi and then we'll have the theta divided by R squared. Then we have the find theta multiplied by the Jihad. Now, a lot of things um variables are actually constant. The Q and the R are both constant, the radius isn't changing. And the charge, the total charge is also not changing. I can pull all those constants out. And so I have E it's equal to, we have the 1/4 pi epsilon knot that's gonna be multiplied by Q divided by pi R squared. And then we'll have the integral of fine data D theta. That's what going in the J hat direction. Now, here my limits of integration, we're gonna go from zero to pi so I can do that integration. So I'll have E is equal to one divided by four pi epsilon knot multiplied by U divided by pi R squared. And the integral of sine is negative cosine, they will have negative cosine of theta that's gonna be evaluated at zero and pi and this still gonna be multiplied by the J hat. So now we can just plug in the limits of integration. So we'll have E is equal to one divided by four pi epsilon knot multiplied by Q divided by pi R squared. And here I will have the quantity of negative cosine pi plus cosine of zero. That's all multiplied by J hat. Well, cosine of pi is just negative one. So it gives me one, then I'll have plus the cosine zeros one that gives me two. So that's just gonna pick up a factor of two. We'll have E equal to one divided by or pi epsilon knot multiplied by two Q divided by pi R squared multiplied by J hat. So that's gonna be my final answer. And if I look at the answer choices that this corresponds to this corresponds to answer D. So just a quick little recap here, we used our definition for the electric field for a continuous charge distribution. But we also use the geometry and the symmetry of the problem in order to eliminate the X component of the electric field, which left only the Y component in the J hat direction. The other thing that we needed to figure out is what are um small charge. Element DQ was actually in terms of quantities that were given. And so we actually had to determine that and it came out to be the Q divided by pi multiplied by D theta. And that's ultimately what we integrated over was the angle theta and that gave us our final expression for the electric field. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Electric Field

Continuous Charge Distribution

Arc Length and Angle Relationship

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