Electric Field due to two circular line charges

[emmett92k]

Homework Statement

Two circular lines of charge are centred at the origin and lie on the xy plane. The inner loop has a radius of a and a total positive charge q. The outer loop has a radius of b and total negative charge -q.

(a) Use Coulomb's Law to calculate the electric field on the z-axis.
(b) Calculate the electric potential of the loops from first principles along the z-axis.
(c) Use the gradient of the electric potential in (b) to calculate the electric field.
(d) Consider a single circular line of charge of radius a, that is split to have a total charge q in the positive y direction, and -q in the negative y direction. Calculate the electric field on the z-axis.

Homework Equations

E = \frac{1}{4\pi\epsilon}\ \int_{-q}^q \frac{\sigma\,dr(2{\pi}rz)}{(z^2 + r^2)^\frac{3}{2}}

V = \frac{1}{4\pi\epsilon}\ \int_{-q}^q \frac{\sigma\,dr(2{\pi}r)}{\sqrt{z^2 + r^2}}

The Attempt at a Solution

(a) I just used the first formula for E

(b) I was confused about the first principles bit so I just used the formulae for V

(c) I differentiated my (b) answer to work back to a similar answer as my (a) answer.

(d) This is where I got stumped a bit. It seems as if it should be straight forward compared to the rest but I can't grasp it.

I would appreciate it if anybody could tell me if my method for all the parts is correct and if they are, if you could point me in the right direction for (d) that would be great. I'm new to Latex so that's why I didn't type up all my answers again. It took me a while to get the first two formulae in. Thanks in advance for any help.

 

[blue_leaf77]

Your equation for E field is not correct, check again the integration limit and variable.

 

[emmett92k]

Hey blue_leaf, should me limit be from a to b and my variable be the space between the discs so b-a?

 

[blue_leaf77]

No, both rings should be treated separately. The way you calculate the electric field due to arbitrary charge distribution is to assume that this charge distribution can be divided infinitely so that you get an infinitesimal piece of charge, since it's infinitesimal you can regard it as a point charge. The electric field from the original charge distribution is then the sum of electric fields from these individual point charges - now you should know the formula for E field due to a point charge.
I'm also wondering where you found that false expression for E field.

 

[Noctisdark]

I agree with blue_leaf, the net electric field is the sum of both electric field due to each of the charged ring, ie $|\vec E| = \frac{1}{4\pi\epsilon_0}\int \frac{z\lambda dArc}{(R^2 + z^2)^{\frac{3}{2}}}$ It's clear that $dArc = Rd\theta$ and you integrate from 0 to 2π, z in the height of the point and R can be either a or b, I'll leave the integration to you,Similary V can be calculated this way !

 

[emmett92k]

Hey Noctisdark, when you say λ is that the same as σ?

 

[emmett92k]

So for (a) I got :\frac{z\sigma}{2\epsilon}\ \Bigg[\frac{a}{(z^2 + a^2)^\frac{3}{2}} + \frac{b}{(z^2 + b^2)^\frac{3}{2}}\Bigg]

 

[blue_leaf77]

So for (a) I got :\frac{z\sigma}{2\epsilon}\ \Bigg[\frac{a}{(z^2 + a^2)^\frac{3}{2}} + \frac{b}{(z^2 + b^2)^\frac{3}{2}}\Bigg]

What you have is a line charge bent to form a circle, so if you follow the common notation, instead of $\sigma$ you should use $\lambda$. Also remember that both rings have opposite charge sign.

 

[emmett92k]

Ok does that mean in the middle there should be a minus sign instead of a plus?

 

[blue_leaf77]

Sorry I forgot to notice another mistake. The two rings have the same magnitude of charge, which is q, and since they have different size, their respective charge density must be different, $\lambda_1 \neq \lambda_2$. So I suggest that rather than expressing the field in term of charge density, you better express it in term of their charges.

Ok does that mean in the middle there should be a minus sign instead of a plus?

It depends on how you determine the axis in the first place, either way the two terms should be of opposite sign.

 

[emmett92k]

So would λ be:

λ = \frac{Q}{2{\pi}r}

 

[blue_leaf77]

Yes.

 

[emmett92k]

So now I'm getting:

E = \frac{zQ}{4\pi\epsilon}\ \Bigg[\frac{1}{a(z^2 + a^2)^\frac{3}{2}} + \frac{1}{b(z^2 + b^2)^\frac{3}{2}}\Bigg]

The a and b outside the brackets seem out of place.

(b) When I go to compute V I use the formula:

V = \frac{1}{4\pi\epsilon}\ \int_a^b \frac{QdR}{\sqrt{z^2 + R^2}}

Is this the correct V formula?

 

[blue_leaf77]

So now I'm getting:

E = \frac{zQ}{4\pi\epsilon}\ \Bigg[\frac{1}{a(z^2 + a^2)^\frac{3}{2}} + \frac{1}{b(z^2 + b^2)^\frac{3}{2}}\Bigg]

The a and b outside the brackets seem out of place.

Yeah, then why are they there?

 

[emmett92k]

Here's my workings:

E = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{z{\lambda}d\theta}{(z^2 + R^2)^\frac{3}{2}}

Subbing in for λ:

E = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{zQd\theta}{2{\pi}R(z^2 + R^2)^\frac{3}{2}}

E = \frac{zQ}{8\pi^2\epsilon}\ . \frac{1}{R(z^2 + R^2)^\frac{3}{2}} . (2\pi - 0)

Subbing my a and b in for R:

\frac{zQ}{4\pi\epsilon}\ \Bigg[\frac{1}{a(z^2 + a^2)^\frac{3}{2}} + \frac{1}{b(z^2 + b^2)^\frac{3}{2}}\Bigg]

 

[blue_leaf77]

Here's my workings:

E = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{z{\lambda}d\theta}{(z^2 + R^2)^\frac{3}{2}}

This is wrong, see comment #5 for the correct one.

 

[emmett92k]

Ah I see so there's an R on top of the fraction so they would cancel. Thank you. And how about my potential formula?

 

[blue_leaf77]

For potential, also calculate it separately for the two rings, using the same argument as before by dividing the rings into infinitesimal point charges.

 

[emmett92k]

But is the forumla I give in comment 13 correct?

 

[blue_leaf77]

Surely not, there you seemed to assume that there are charge distribution in between the rings, while in fact there are not. As I said, the potential is also calculated separately for the two rings.

 

[emmett92k]

I'm assuming I would calculate it in some variance of the following formula:

V = \frac{1}{4\pi\epsilon}\ \int \frac{Q}{R}

However should there be a 'z' in there somewhere and are the limits 0 to 2π with a dθ on the top too?

 

[blue_leaf77]

That $Q$ should be $dq$.

However should there be a 'z' in there somewhere and are the limits 0 to 2π with a dθ on the top too?

Why do you think there should be z? Wouldn't that then change the physical meaning of the total potential being the sum of potential due to infinitesimal point charges?
To determine the integration limits, use the same argument before. You are given a line charge distribution in a form of circle, how can you change $dq$ to accommodate an arc element as before.

 

[emmett92k]

I was thinking there should be a z because in part (c) I will have to find the electric field from the gradient which is just partial differentiation. I thought this electric field would be the same as the electric field in part (a) which does have a z.
So are the limits 0 to 2π again giving me:

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

 

[blue_leaf77]

I thought this electric field would be the same as the electric field in part (a) which does have a z.

This time we are calculating potential, which is a scalar quantity, while in part a) we calculated E field which is a vector. The z in the integrand in part a) comes from the fact that E field is a vector, since the observation point lies in the axis, the transversal vector components from the different point on the ring cancel out leaving us with the longitudinal component only. To calculate the longitudinal component you must multiply the E field magnitude with cosine the angle subtended by the observation point and the ring. It's this cosine which gives z in the integrand.

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

The total potential reads as
$$ V = \frac{1}{4\pi\epsilon}\ \int \frac{dq}{r} $$
Now make substitution for $dq$ as before. Keep in mind that $r$ in the denominator is the distance of a given point on the ring to the observation point, not the ring's radius.

 

[emmett92k]

So:

V = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{Rd\theta}{r}

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

 

[blue_leaf77]

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}

 

[blue_leaf77]

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]

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]

$dq$ 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]

Yes that's the correct answer.

 

[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]

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.

 

[emmett92k]

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]

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.

 

[emmett92k]

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]

You are being careless in calculating the derivatives.

 

[emmett92k]

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]

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.

 

[emmett92k]

So does that mean I split the circle into 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]

So does that mean I split the circle into 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.

 

[emmett92k]

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

 

[blue_leaf77]

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.

 

[emmett92k]

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?

 

[emmett92k]

Actually is it:

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

 

[blue_leaf77]

Not that one, look at the link I just added in my previous comment. What I meant is the expression of the unit vector $\hat{r}$, not the coordinates themself.

 

[emmett92k]

Did you see I left two comments there? Because my first comment is what is on that page?

Do you mean $\hat{r}=\frac{\vec{r}}{r}$