Electric potential on circle problem

[physicsjock]

[PLAIN]http://img526.imageshack.us/img526/5079/29693720.jpg
im struggling to start this question
do i calculate all of the potentials seperatly and add them up?
and to calculate the potential on the circle and 1/4 circle would i just use 2pi r ?
and for the 1/4 cirlce definite intergrate it from pi/2 to pi
how would i go about the diapole ?

 

[AEM]

Yes, you calculate each potential separately and then add them all up. For the curved pieces construct an integral for a potential dV from a bit of charge dq. then integrate. Your limits of integration depend on how you choose to draw your diagram, but the total angle is the same no matter how you set it up. For the dipole, do each charge separately.

 

[physicsjock]

would i be able to have a hand with the intergral
im kind of struggling to work it out
so the potential from one point on the big circle is k(-3x10^-6 / 0.06)
where k = 8.99x10^9
so would i times this by the intergral of 2pir?
but that just turns into the area of a circle
coudl i just times that by the perimeter or the circle?
and for the smaller curve could i times its charge by 1/4 2pi r

 

[CanIExplore]

Let's start with the potential due to a point charge.

$$\begin{flalign*} V & = & & k \ \frac{q}{r} \ \mbox{ or } \ \frac{1}{4 \pi \epsilon_0} \ \frac{q}{r}\\ \end{flalign*}$$

We want to break the ring into infinitesimal segments and find the potential due to one of these segments right? Then we can add them all up using the integral. So we want to calculate V=$$\int dV$$. What is the equation for dV?

 

[Daniiel]

k * intergral (dq/r)
is that right?
how do i relate it to the perimeter of the circle? cause it doesn't say how thick the wire is

 

[CanIExplore]

k * intergral (dq/r)
is that right?
how do i relate it to the perimeter of the circle? cause it doesn't say how thick the wire is

Yes that's right. Now how else you can express dq? There is a total charge Q spread uniformly on the ring and you know the size of the ring, since you are given the radius.

When a problem states an object is "thin" they mean it has no thickness. So imagine the ring to be a line bent into a circle of radius R.

 

[Daniiel]

ok so is the total charge Q = -3 microC ?
so it woudl be like

k* (Q(2pir)dr)/root(z^2 + r^2) ?
thats what they get in the book, but I am just abit confused
why is it z^2 + r^2? it doesn't make sense with phythagorus unless I am thinking silly

 

[CanIExplore]

ok so is the total charge Q = -3 microC ?
so it woudl be like

k* (Q(2pir)dr)/root(z^2 + r^2) ?
thats what they get in the book, but I am just abit confused
why is it z^2 + r^2? it doesn't make sense with phythagorus unless I am thinking silly

Let's start by defining our coordinate system. The simplest case would be to place the ring in the xy plane, and center it on the z-axis. Now if we do this, then according to our diagram we are evaluating the potential at the origin of our coordinate system correct?
So in our new equation for the potential,
$$\begin{flalign*} dV & = & & \ \frac{1}{4 \pi \epsilon_0} \ \frac{dq}{r}\\ \end{flalign*}$$, r is the distance from the differential segment to the point where we are calculating the potential so, in our case this is the radius since we want to evaluate the potential at the origin. Now think about how we can express dq. We want to integrate over all the segments of the ring right? So we need for dq to somehow be expressed in terms of some differential length. We know the total charge on the ring and we know how long it is so we can define a linear charge density as $$\lambda = Q/l$$ where l is the length of the ring and thus has units of charge/length. Then dq is?

 

[Daniiel]

ohh i see
so dq is lamda dx
then because its a constant you can pull it out of the intergral?
and it looks like

k lambda (ingtergral of(dx/r))
is that right?
the next step puts r = root(x^2 + d^2)
is r always = to somthing like that?
is that like a rule I've missed?

 

[CanIExplore]

ohh i see
so dq is lamda dx
then because its a constant you can pull it out of the intergral?
and it looks like

k lambda (ingtergral of(dx/r))
is that right?

That's correct, although I would write it as dl, so you don't get confused with integrating over the x coordinate. So this is just the line integral around the loop which is just equal to the perimeter.

k* (Q(2pir)dr)/root(z^2 + r^2) ?
thats what they get in the book, but I am just abit confused
why is it z^2 + r^2? it doesn't make sense with phythagorus unless I am thinking silly

This expression for r is for the more general case. Suppose we have the ring lying in the xy plane, centered on the z-axis as before, but now we want to evaluate the potential due to the ring at some point on the z-axis a distance z above the origin. Remember r in our equation for the potential is the distance from the element dq to the point at which we're evaluating the potential. Then r is no longer just equal the radius of the ring since we're not finding the potential at the origin, instead it is the distance from the element dq to our point of evaluation somewhere on the z-axis above the ring. So if you draw this out, you can use the pythagorean theorem to find the distance r. Do you see how they get sqrt(z^2 + R^2)? To avoid confusion I am using R as the radius of the ring. r is still the distance from dq to the point where we are evaluating the potential.

 

[Daniiel]

ohhh right right right
i get it now
sorry i was reading two different examples
one was a disc and the other was a line
the disc one was d
but i udnerstand now

so now i juts intergrate with the limits L and 0 right?
and thatll give me the charge for the cirlce?

ok i think i got that part down, and with the next part, i juts do somthing similar with the length of the curve and a new distance?
so i'd add them together?
and ill just have a look at how to work with the dipole

 

[Daniiel]

sorry i have to go out for a bit
thanks for your help so far its been really good
im just looking at this disc example
how come we wouldn't use
dq = sigma 2pir dr
cause the line example looks like its finding the charge on a line that gets further and further away from the point
but the ring stays the same distance from the point
does that matter?

 

[Daniiel]

aw man
i intenesly over think these questions
but I've been thinking
and this is what made me think what i said before
like because it looks like by this pictre the intergral is finding the charges at different distances from the point
and the ring has a constant distance from the point
i dono i hope my blabber makes some kind of sense
thanks for bearing with me
[PLAIN]http://img168.imageshack.us/img168/9555/26440124.jpg
does it add it up with a different r every time? like a different distance from the point?

 

[CanIExplore]

im just looking at this disc example
how come we wouldn't use
dq = sigma 2pir dr

That is what you would use. Why do you think you wouldn't be able to write dq this way?

like because it looks like by this pictre the intergral is finding the charges at different distances from the point
and the ring has a constant distance from the point
does it add it up with a different r every time? like a different distance from the point?

You're right, we are finding the potential due to each infinitesimal charge dq and then adding them all up. What we end up doing is expressing the charge as dq= $$\lambda dx$$ so that we end up integrating over the coordinate x, or in the case of a surface charge density the area da, etc.. This is allows us to account for the geometry of the problem. In our equation $$\begin{flalign*} dV & = & & \ \frac{1}{4 \pi \epsilon_0} \ \frac{dq}{r}\\ \end{flalign*}$$ for the potential, r is the distance from the point dq to the point at which we're evaluating the potential. So in the ring problem, this is just equal to R, the radius of the ring since we were evaluating the potential at the origin. Every dq is the same distance R from the origin so we just take r=R=constant and pull it out of the integral. Now in the case of the disc, r is now equal to sqrt(x$$^{2}$$ + d$$^{2}$$) as you can see from the diagram. And we're integrating over dx, so we cannot just pull r out as a constant in the same way we did for the ring problem. Since r is a function of x, it remains inside the integral.

 

[Daniiel]

so would we end up getting as the intergral

(k lamda)/r Intergral(dl)
r as a constant and out of the intergral?

so it would end up being L * (k lamda)/r + 0 ? with limits L and 0

 

[CanIExplore]

so would we end up getting as the intergral

(k lamda)/r Intergral(dl)
r as a constant and out of the intergral?

so it would end up being L * (k lamda)/r + 0 ? with limits L and 0

For the line charge pictured in the diagram that you uploaded r is not constant. Remember r is the distance from dq to the point at which we're evaluating the potential. So for the ring, r is constant and equal to the radius. For the disc, r is not constant. And for the line charge in that diagram r is most certainly not constant. As you move down the line from the origin to L, you're moving further and further away from the point of evaluation, r is changing the whole time. r as shown in the diagram is r=sqrt(x$$^{2}$$ + d$$^{2}$$) which is a function of x. And dq is dq=$$\lambda dx$$ so we're integrating over x. So like I said before r cannot be pulled out of the integral since it is a function of x. Look again at the diagram. That is the equation for dV, that is what you integrate to obtain the potential at P due to the line segment, its an integral over x.

 

[physicsjock]

so for the ring r is constant
so we don't treat it the same as that example of the line, we pull the r outside the intergral and intergrate?

 

[CanIExplore]

so for the ring r is constant
so we don't treat it the same as that example of the line, we pull the r outside the intergral and intergrate?

Correct, r is constant for the ring so you can pull it out of the integral. The integral is then just a line integral over the length of the ring which is just its perimeter.

 

[physicsjock]

oh sweet
so it would look somthing likkee
krQ * (intergral(dr/2pir))
with limits 2pi and 0?
did i stuff up the Q? lambda = Q/l right? and Q is constant so i pulled it out

 

[Daniiel]

would the angle between the dipole be 180 degrees or 90?
if its 90 then its just 0
but then if its 180 V will be negative
and we don't know r

V = k x (pcos(theta))/r^2)
where p is the dipole moment