Question

Alright so let me check If I have this right...Electric Field of Semicircle...

Answer 1

Find the electric field due to a half circle of radius R and total charge Q. (a) First estimate by dividing the charge into five equal parts. (b) Repeat for 9 equal parts, (c) Using integral calculus solve exactly.

So we would have



a) Break into 5 equal pieces

Answer 2

Total Charge = Q

So charge here is Q/5

Charge per unit length =\(\large \frac{Q}{L}\)

Q will be 1/5Q and L would be half the circle divided into 1/5 parts...so... \(\large \frac{1}{5} \frac{(1/5)Q}{\pi r} = \frac{(1/5)Q}{5 \pi r}\)

Answer 3

Let that = \(\large \lambda\)

So
\[\large \lambda = \frac{.2Q}{5\pi r}\]

Now charge on each slice would be

\[\large dQ = \lambda Rd\theta\]

So we would have a small portion of the electric field as

\[\large dE = \frac{kdQ}{r^2} = \frac{k\lambda }{r}d\theta\]

Answer 4

Components of each slice are

\[\large dE_x = dEcos \theta\]
\[\large dE_y = dEsin\theta\]

So \(\large E_x\) would be

\[\large \int^{\pi}_{0} \frac{k \lambda}{r}cos \theta d\theta = \frac{k \lambda}{r} \int^{\pi}_{0} cos\theta d\theta = \frac{k \lambda}{r}sin\theta \left.\right|_0^\pi\]

And \(\large E_y\) would be

\[\large \int^{\pi}_{0} \frac{k \lambda}{r}sin \theta d\theta = \frac{k \lambda}{r} \int^{\pi}_{0} sin\theta d\theta = -\frac{k \lambda}{r}cos\theta \left.\right|_0^\pi\]

Right? and then just solve knowing \(\large \lambda = \frac{.2Q}{5\pi r}\)

Answer 5

And for 'b' just replace that 5 on the denominator with 9 instead...and finally integration just using \(\large \pi r\)as the denominator since we want it all?