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I am stuck with this problem:

A semicircular wire with radius R has uniform charge density−λ. Show that at all points along the “axis” of the semicircle, the vectors of the electric field all point toward a common point in the plane of the semicircle. Where is this point?

enter image description here

What I have so far is:

$-\lambda=\frac{dq}{ds}$

$\overrightarrow{E}=\frac{1}{4\pi \varepsilon_o}\int \frac{-\lambda ds}{r^2}\widehat{r} = \frac{-\lambda}{4\pi \varepsilon_o}\int \frac{Rd \phi}{r^2}\widehat{r} $

Maybe it is a good idea to use cylindrical coordinates, so that $\widehat{r}=\langle \widehat{\rho},\widehat{\theta},\widehat{z} \rangle$. But I am not really sure how to use this to solve this problem.

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You are almost there:

$r^2=R^2+z^2$. In terms of a coordinate system with z vertical and x along the symmetry axis of the semicircle, you can write the coordinate of the point where you measure the electric field as $(0,0,z)$ and the source (an element of charge $dq=-\lambda R d\theta$) is at $(R\cos\theta,R\sin\theta,0)$. Then $\vec{r}=(0,0,z)-(R\cos\theta,R\sin\theta,0)=-(R\cos\theta,R\sin\theta,-z)$, and $\hat{r}=-(R^2+z^2)^{1/2}(R\cos\theta,R\sin\theta,-z)$. You will have three components for $\vec{E}$, given by $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{\lambda R}{(R^2+z^2)^{5/2}}\int_{-\pi/2}^{\pi/2}d\theta(R\cos\theta,R\sin\theta,-z)$$ You can now solve the three integrals (for the $x,y,z$ components). A sanity check is that the integral along $y$ is 0. If all $\vec{E}(z)$ are directed to a point in the plane, along the $x$ axis, say at a distance a from the center, then $$-\frac{E_z}{E_x}=\frac{z}{a}$$ All you need is now to find a.

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  • $\begingroup$ Thanks, but $\widehat{r}=-(R^2+z^2)^{-1/2}(R\cos \theta , R \sin \theta , -z)$ right? And also in the third integral $z$ can be taken out of it because it doesn't depend on $\theta$? $\endgroup$ – math4everyone Aug 28 '16 at 17:20
  • $\begingroup$ Correct in both respects. You notice that the term before the integral is common to all components, so it will cancel in the last equation $\endgroup$ – Andrei Aug 28 '16 at 18:45
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Since this is basically a homework type question, I will not give the full solution. The diagrams are given: enter image description here

Consider a charge element having charge $y$$dx$ where y is the charge density, and x is the angle shown in TOP VIEW. From figure 2, you can see that at a point A at a height z, The electric field is quite easily found using the square of the radial distance $AC^2$ = $AB^2$ = ($r^2 + z^2$) [ r is the radius]. Now , You need to find its components. So multiply this field you found by sin($tan^-1$(z/r)) to obtain the vertical component of the field at A. Similarly, multiply the field by $cos(tan^-1(z/r))$ to obtain the field magnitude in the direction OB and OC. Multiply by cos(x) to obtain the field components in the X direction.[note that the y direction field components cancel out]. Now you have the vector components in terms of differential charge. Integrate it to obtain the field at A due to the the ring.(trivial integration). Now that you have the vector, i will leave the rest to you to determine where that vector cuts the X axis. [Hint: use your knowledge of lines from coordinate geometry, or any other way you want]. Let me know your answer.

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