Expression for the magnitude of an electric field

[mousey]

Homework Statement

Find an expression for the magnitude of the electric field at point A midway between the two rings of radius R shown in the figure below. The ring on the left has a uniform charge q1 and the ring on the right has a uniform charge q2. The rings are separated by distance d. Assume the positive x axis points to the right, through the center of the rings. (Use any variable or symbol stated above along with the following as necessary: k for Coulomb's constant. Assume q1 is greater than q2, and that both charges are positive.)
(picture in replies)

Relevant Equations

E= (kq)/d^2

k(q_1-q_2)/(R^2+x^2)^3/2

 

[mousey]

 

[Delta2]

You basically, have to consider one ring, let's say the left one, and calculate the magnitude of the electric field at A due to it, that is calculate the line integral $$\vec{E_1}=\int_{ring 1}\frac{k\rho_1 (\vec{r}-\vec{r'})dr'}{|\vec{r}-\vec{r'}|^3}$$, where $$\vec{r}=\frac{d}{2}\hat x$$,$$\rho_1=\frac{q_1}{2\pi R}$$, $$dr'=\sqrt{1+(f'(z))^2}dz$$ $$f(z)=\sqrt{R^2-z^2}$$, $$\vec{r'}=\sqrt{R^2-z^2}\hat y+z\hat z$$.

It is enough to calculate only the x-component of $\vec{E_1}$, because the point A lies on the axis that passes through the center of the ring, hence due to symmetry reasons, the y and z components of $\vec{E_1}$ will be zero at point A.

Then calculate similarly the contribution $\vec{E_2}$ to the electric field from the ring to the right, and then take the sum $$\vec{E}=\vec{E_1}+\vec{E_2}$$ for the total electric field E at point A.

 

[mousey]

Thank you. I'm just wondering, the problem says q1 is greater than q2, and both are positive, so wouldn't I subtract E1-E2? Do I integrate both and then subtract?

 

[haruspex]

Thank you. I'm just wondering, the problem says q1 is greater than q2, and both are positive, so wouldn't I subtract E1-E2? Do I integrate both and then subtract?

The fields add as vectors, snd as scalars along the x axis. But as scalars, one will have a negative value.
Yes, find each field separately before combining them.

 

[Delta2]

Yes , well if you find that $$\vec{E_1}=A\hat x$$ where A is a scalar function of $q_1,R,d$ then you ll find that $$\vec{E_2}=-B \hat x$$ where B is another scalar function of $q_2,R,d$, so that $$\vec{E_1}+\vec{E_2}=(A-B)\hat x$$

Just remember that when you do all the substitutions (for $\vec{r},\vec{r'} , dr$ e.t.c) you ll end up with an integral that has $\hat y$ and $\hat z$ terms as well as the $\hat x$ term. Just ignore the first two terms (or maybe you can integrate as an excercise to prove that those integrals would be zero) and focus on the $\hat x$ term that is the x-component of the E-field.

And actually the way I 've setup $\vec{r'}$ is for the integral of the upper half of the ring. To calculate the integral for the lower half you got to set $\vec{r'}=-\sqrt{R^2-z^2}\hat y+z\hat z$. This also means that when you calculate the x-component you got to multiply it by 2 (the x-component is the same for the upper half and for the lower half).