A uniformly charged insulating rod of length 13.0 cm is bent...

A uniformly charged insulating rod of length 13.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of -7.50 µC. (a) Find the magnitude of the electric field (in N/C) at O, the center of the semicircle. 283000 \times N/C (b) Find the direction of the electric field at O, the center of the semicircle. to the left to the right upward downward into the page out of the page (c) What if? What would be the magnitude of the electric field (In N/C) at O if the top half of the semicircle carried a total charge of -7.50 µC and the bottom half, insulated from charge of +7.50 µC? 283000 \times N/C

Transcript

00:01 Okay, so for this question, so the electric view should be equal to kq over r squared and sine fidd, which is the horizontal components of the electric view, okay? it's because the vertical components will be eventually canceled out to do symmetry.

00:16 So we have a horizontal component need to be considered.

00:19 So when a lambda, we see linear charge density, or you can say charge per unit length, is equal to q over l.

00:25 Then we have q is equal to landa l.

00:29 And then we have dq is equal to lambda times d.

00:33 Then the de is equal to kdq over r squared times sine theta, which we have k lambda times dl over r square and sine theta.

00:40 And we also know that dl is equal to r d theta is because l is equal to 2 pi x theta or 2 pi, which is r theta.

00:48 So if we put d on each side, then we have dl is equal to r d theta.

00:53 Then we're plugging this equation here to the next step.

00:58 Then we'll have d is equal to k lambda r d theta over r square times sine theta.

01:03 Then we have e is equal to the integral from 0 to pi, which is k landa r over r square sine theta over inside, and then the d theta on the right side.

01:14 And eventually we have integral 02 pi and then k landa over r times sine theta and then times d theta.

01:22 And then let's start our calculation.

01:25 Then we have k landa over r times 02 pi, sine theta, d theta, which, which is a which will give us k lambda over r times the antidearity of sine theta is negative cosine theta, okay? and the range is from 0 to pi.

01:54 So therefore we have k landa over r times negative cosine pi minus negative cosine zero and this will give us k -landa over r times 1 plus 1, which will give us e is equal to 2k landa over r.

02:27 Okay? remember lambda is equal to q over l so therefore we can plug in this equation here back into the equation here and we should have e is equal to 2k times q over l over r...

Question

Please give Ace some feedback