# Electric Field due to two circular line charges

blue_leaf77
Homework Helper
How did you arrive in the second line? The first line is just the potential due to a single ring.

emmett92k
This is how:

$V = \frac{R}{4r\pi\epsilon}\ \int_0^{2\pi} d\theta$

Then I subbed my a and b in for R separately. With b being negative because of the negative charge, then I summed them to get:

$V = \frac{1}{2\epsilon}\ \frac{a-b}{r}$

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blue_leaf77
Homework Helper
As I have pointed out, ##r## is the distance from the observation point to a given point on the ring charge distribution. Which means obviously ##r## for the small and big rings are not the same, ##r_1 \neq r_2##. Ok to proceed in an easier way, for now just remove the bigger ring from your mind and calculate the potential due to the small ring only and let me know the result.

emmett92k
So $V_1 = \frac{1}{2\epsilon}\ \frac{a}{r_1}$

blue_leaf77
Homework Helper
So $V_1 = \frac{1}{2\epsilon}\ \frac{a}{r_1}$
There are still many missing quantities in your equation. I suggest that you start from the integral in comment #24. Then use the fact that ##r## is a constant (do you know why?), hence it can be taken out of the integral leaving only ##\int dq##, what is then this integral equal to?

emmett92k
Is the integral of ##dq## not the integral of $Rd\theta$?

blue_leaf77
Homework Helper
##dq## is equal to ##\lambda R d\theta##.

emmett92k
Ah so I end up with $V_1 = \frac{Q}{4{\pi}r_1\epsilon}$, $V_2 = \frac{-Q}{4{\pi}r_2\epsilon}$

This give a total V of:

$V = \frac{Q}{4\pi\epsilon}\Bigg[\frac{1}{r_1} - \frac{1}{r_2}\Bigg]$

Is this correct?

blue_leaf77
Homework Helper

emmett92k
Ok thanks for all the help by the way.

For part (c) I know its partial differentiation but what will I differentiate with respect to? Will I do $r_1$ and $r_2$ separately?

blue_leaf77
Homework Helper
Ok thanks for all the help by the way.
You are welcome.
For part (c) I know its partial differentiation but what will I differentiate with respect to? Will I do $r_1$ and $r_2$ separately?
Part c) is a bit tricky because as you said we need to apply nabla operator ##\nabla## to the potential at arbitrary point. But what we have at hand is the potential at points along the axis only. The trick is to make use of the symmetry of the system. So, first start by writing ##V(x,y,z)## as the potential at any point in space, not just on the axis. We want to calculate
$$\mathbf{E}(x=0,y=0,z) = -\nabla V(x,y,z) \big|_{x=0,y=0} = - \bigg( \hat{x}\frac{\partial V(x,y,z)}{\partial x}|_{x=0,y=0} + \hat{y}\frac{\partial V(x,y,z)}{\partial y}|_{x=0,y=0} + \hat{z}\frac{\partial V(x,y,z)}{\partial z}|_{x=0,y=0} \bigg)$$
Where the requirement x=0 and y=0 indicates that the observation point is on the axis which is taken to be the z axis. Now apply some symmetry argument to find the value of the first two terms without really calculating it.

Well the first two terms would be equal to zero. How do I differentiate the ##z## part if there is no ##z## variable?

blue_leaf77
Homework Helper
That's right but I hope you know why they must be zero. To reveal the position of z, you need to express ##r_1## in term of the corresponding ring radius and the distance from the ring center to the observation point, which is z. Do the same for the other ring.

Ok I'm starting to follow now. The r values will be the distance of the hypotenuse if I draw a triangle. So:

$r_1 = \sqrt{a^2 + z^2}$, $r_2 = \sqrt{b^2 + z^2}$

So I sub them and differentiate to get:

$E = \frac{Q}{4\pi\epsilon}\Bigg[\frac{2z}{(a^2+z^2)^\frac{3}{2}} - \frac{2z}{(b^2+z^2)^\frac{3}{2}}\Bigg]$

This simplifies to:

$E = \frac{Qz}{2\pi\epsilon}\Bigg[\frac{1}{(a^2+z^2)^\frac{3}{2}} - \frac{1}{(b^2+z^2)^\frac{3}{2}}\Bigg]$

This is different to my E in part (a) however.

blue_leaf77
Homework Helper
You are being careless in calculating the derivatives.

Sorry I've found the mistake, the 2 multiplies by a half and cancels.

As regards part (d) do I treat it as a dipole and use the formula:

$E(r,\theta) = \frac{qd}{4{\pi}{\epsilon}r^3}(cos{\theta}\hat{r}+sin\dot{\theta}\hat{\theta})$

blue_leaf77
Homework Helper
I don't know if that works. But anyway, task d) is not so difficult compared to task a). The way you do it is the same as we did in a), only that now there is only one ring but with upper half positively charged and lower half negatively charged. Then make use of the symmetry property to deduce which vector components of the E field should vanish.

So does that mean I split the circle in to two semi-circles and use the same method as (a) but change $\lambda = \frac{Q}{2{\pi}R}$ to $\lambda = \frac{Q}{{\pi}R}$ because its half the circumference?

blue_leaf77
Homework Helper
So does that mean I split the circle in to two semi-circles and use the same method as (a) but change $\lambda = \frac{Q}{2{\pi}R}$ to $\lambda = \frac{Q}{{\pi}R}$ because its half the circumference?
Yes but that will come later.
For this problem it's easier if we work in spherical coordinate and translate the observation point to the origin, so that the ring will lie on a plane located at ##z##. I will start with the E field due to the positive charge first
$$\mathbf{E}_+ = k\int_0^\pi \frac{\lambda R d\phi}{r}\hat{r}$$
Next, express ##\hat{r}## in its cartesian components.

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For ##E_+## is ##\hat{r}## r(0,a,z)?

blue_leaf77
Homework Helper
For ##E_+## is ##\hat{r}## r(0,a,z)?
Nope, ##\hat{r}## is the unit vector from a particular point on the ring to the origin, which means it's also the same as the inverse of the radial unit vector in spherical coordinate. So you will want to know how to express radial unit vector in spherical coordinate into cartesian components, ##\hat{x}##, ##\hat{y}##, and ##\hat{z}##.
You can find what you are looking for in https://en.wikipedia.org/wiki/Spherical_coordinate_system under "Integration and differentiation in spherical coordinates" section.

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Spherical coordinates are ##(r,\phi,\theta)## and cartesian is ##(x,y,z)##.

##r=\sqrt{x^2+y^2+z^2}##, ##\theta=tan^{-1}\big(\frac{y}{x}\big)##, ##\phi=cos^{-1}\big(\frac{z}{r}\big)##

How do I precede from here?

Actually is it:

##x=rcos{\theta}sin\phi##
##y=rsin{\theta}cos\phi##
##z=rcos\phi##

blue_leaf77