Is this a correct mathematical solution?

[transgalactic]

is this a correct mathematical solution??

question:
http://i40.tinypic.com/2hwcswm.gif

solution:
$$\sigma=\frac{7.5\mu c}{14}=0.53571428571428571428571428571429\mu c\\$$
$$14=2\pi r\\$$
$$r=\frac{14}{\pi}\\$$
$$dq=\sigma r d\theta \\$$
$$k\sigma r\int_{0}^{\pi}d\theta=k\sigma r (\pi-0)=...$$

is it a correct solution?

 

[dx]

Hi transgalactic,

The rod is bent into a semicircle, not a circle, so you should have 14 = (2πr)/2 = πr. What's your reasoning behind the integral at the end?

 

[transgalactic]

ohh you are correct
but thus is a minor mistake.
the main issue is with the integral.
i took a small part of a charge and calculated its contribution on the field of point p.
and then i sum them..

is it correct

 

[tiny-tim]

Hi transgalactic!

Your charge density σ = 7.5/14 µC and your radius r = 14/π are correct …

but I don't understand what τ is, nor your dq = στdθ equation …

you need to integrate the field (a vector) from each dθ segment …

what is it?

EDIT: oops … I thought your r was a τ … sorry

 

[dx]

No, it doesn't look correct to me.

Let's do it step by step. The small charge within dθ is (σ⋅r⋅dθ). You got that correct. This small charge is a distance r away from the point O. Now use Coulombs law to find the magnitude and direction of the electric field at O due to this small charge.

 

[transgalactic]

i forgot to use the formula of a field.
$$\frac{k\sigma r}{r^2}\int_{0}^{\pi}d\theta=\frac{k\sigma }{r} (\pi-0)=$$
now its correct?

 

[dx]

You must take into account the fact that the vectors are not all pointing in the same direction. Your integral just gives you the sum of the magnitudes of the E field due to each small charge. That's not what you want. What you want is the vector sum. Draw a picture. What does the symmetry of the situation tell you about the final direction of the field? Using the symmetry, you will be able to ignore one of the components in the vector sum.

 

[transgalactic]

i know that the "y" components are being canceled
i thought that the integral sorts that out.
how to do it in a vectorized way so it will cancel the y components by itself,
and not by the way of taking components of the feils of each part
??

 

[dx]

You used Coulom's law to find the magnitude of the field to to a small charge (σ⋅r⋅dθ) at the point P:

$$|dE| = \frac{r d\theta \sigma}{4\pi\epsilon_{0} r^2}$$.

What are the x and y components of this vector? If the x-component is dE_x, then the x-component of the final vector will be

$$\int dE_{x}$$

Similarly for the y-component. You can tell that the y-component of the final vector will be zero because of the symmetry, so you just have one integral to do.

 

[transgalactic]

i get the x component by multiplying by cos theta
and because of the simitry i need to multiply by 2
$$2cos \theta \frac{k\sigma r}{r^2}\int_{0}^{\pi}d\theta=2cos \theta\frac{k\sigma }{r} (\pi-0)$$
correct?

 

[dx]

Why is cos(θ) outside the integral? It should be inside. And there's no need to multiply by 2.

You found that

$$dE_{x} = \frac{k r \sigma d\theta}{r^2} \cos(\theta)$$.

So x-component of the final vector is

$$\int dE_{x} = \int_{0}^{\pi} \frac{k r \sigma d\theta}{r^2} \cos(\theta)$$

 

[tiny-tim]

And there's no need to multiply by 2.

unless you make the integral from 0 to π/2

 

[transgalactic]

so we take the sum of all x components from 0 to \pi (no multiply by 2)
$$\frac{k\sigma r}{r^2}\int_{0}^{\pi}cos \theta d\theta= ..$$

 

[dx]

unless you make the integral from 0 to π/2

Now why didn't I think of that! You're so clever tiny-tim.

 

[transgalactic]

thanks

 

[dx]

so we take the sum of all x components from 0 to \pi (no multiply by 2)
$$\frac{k\sigma r}{r^2}\int_{0}^{\pi}cos \theta d\theta= ..$$

Yes, that's correct.

(BTW, you're taking k = 1/(4πε₀) right?)

 

[tiny-tim]

erm … with those limits, isn't it ∫ sinθ dθ?

 

[dx]

Oops, sorry! tiny-tim is right. It should be sin.

 

[transgalactic]

sin is the y components
cos is the x

 

[dx]

It depends on where you're measuring theta from.

 

[transgalactic]

look at the photo at the start of the thread
x component is cos theta

 

[dx]

You didn't say where you were measuring theta from in your picture.

http://img36.imageshack.us/img36/9175/moooss.jpg

 

[tiny-tim]

look at the photo at the start of the thread
x component is cos theta

(No, the photo is vague about that. But anyway:)

If you want the x component to be cosθ, you must integrate from -π/2 to π/2

 

[transgalactic]

yes!

 

[transgalactic]

You didn't say where you were measuring theta from in your picture.

http://img36.imageshack.us/img36/9175/moooss.jpg

[/URL]

like that

 

[transgalactic]

(No, the photo is vague about that. But anyway:)

If you want the x component to be cosθ, you must integrate from -π/2 to π/2

ohhh you are correct too
thanks

 

[dx]

If it's like that, then you integrate from 0 to pi, and it will be sin.

If you measure it from the negative x-axis, it will be cos, and the limits will be -pi/2 to pi/2.

 

[transgalactic]

but it Cartesian system on polar coordinates.
i need to draw cartesian x axes and y axes even if it polar coordinates
?

 

[dx]

You can use both at the same time for different parts of the problem, nothing wrong with that.