Doubts about the electric field created by a ring

[Guillem_dlc]

I have the calculation of the electric field created by a ring of radius $R$ uniformly charged with a linear density of charge $\lambda$ at any point on the axis perpendicular to its surface ($z$ axis), but I have some doubts about it. I'll leave you the calculation done first:

In $x$ axis the field is $0$. We calculate the electric field:
$$dE=k\dfrac{dq}{r^2}\rightarrow dq=\lambda dl$$
$$E=k\int_0^L \dfrac{\lambda dl}{r^2}=\dfrac{k\lambda}{r^2}\int_0^L dl=\dfrac{k\lambda}{r^2}2\pi R$$
because we have a ring: $2\pi R$
$$E=\dfrac{1}{4\pi \varepsilon_0}\dfrac{2\pi R}{r^2}=\dfrac{\lambda}{2\varepsilon_0}{R}{r^2}=\dfrac{\lambda R}{2\varepsilon_0 r^2}$$
$$E_z=E\cdot \cos \alpha =E\cdot \dfrac{z}{r}=\dfrac{\lambda}{2\varepsilon_0}\dfrac{zR}{r^3}=\dfrac{\lambda zR}{2\varepsilon_0 (z^2+R^2)^{3/2}}$$
because $r=\sqrt{z^2+R^2}$.

Question: I have problems with these exercises with integrals. For example, I understand this, but the drawing of the principle I don't see because it takes two instead of one and in the integral I don't understand because in some exercises there are things that are constant and things that are not.

 

[PeroK]

I have the calculation of the electric field created by a ring of radius $R$ uniformly charged with a linear density of charge $\lambda$ at any point on the axis perpendicular to its surface ($z$ axis), but I have some doubts about it. I'll leave you the calculation done first:
View attachment 298668

In $x$ axis the field is $0$. We calculate the electric field:
$$dE=k\dfrac{dq}{r^2}\rightarrow dq=\lambda dl$$
$$E=k\int_0^L \dfrac{\lambda dl}{r^2}=\dfrac{k\lambda}{r^2}\int_0^L dl=\dfrac{k\lambda}{r^2}2\pi R$$
because we have a ring: $2\pi R$
$$E=\dfrac{1}{4\pi \varepsilon_0}\dfrac{2\pi R}{r^2}=\dfrac{\lambda}{2\varepsilon_0}{R}{r^2}=\dfrac{\lambda R}{2\varepsilon_0 r^2}$$
$$E_z=E\cdot \cos \alpha =E\cdot \dfrac{z}{r}=\dfrac{\lambda}{2\varepsilon_0}\dfrac{zR}{r^3}=\dfrac{\lambda zR}{2\varepsilon_0 (z^2+R^2)^{3/2}}$$
because $r=\sqrt{z^2+R^2}$.

Question: I have problems with these exercises with integrals. For example, I understand this, but the drawing of the principle I don't see because it takes two instead of one and in the integral I don't understand because in some exercises there are things that are constant and things that are not.

You got the right answer, but there are problems with what you did. First, the electric field is a vector quantity. You must analyse the vector components from the beginning. In this case, the $x$ and $y$ components cancel out by symmetry, leaving only the $z$ component, which is the same for any point on the ring.

That allows you to do this calculation without using integration:
$$dE_z = |d\vec E|\cos \alpha = \frac{1}{4\pi \epsilon_0}\frac{dq}{r^2}\cos \alpha$$
And, as $z$ is fixed, everything is constant, so:
$$E_z = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\cos \alpha$$
And you can express $Q, r$ and $\alpha$ in terms of $\lambda, R$ and $z$ to get your answer.

Of course, you can integrate round ther ring if you want to, but it must give simply the total charge $Q = 2\pi R\lambda$.

 

[Guillem_dlc]

You got the right answer, but there are problems with what you did. First, the electric field is a vector quantity. You must analyse the vector components from the beginning. In this case, the $x$ and $y$ components cancel out by symmetry, leaving only the $z$ component, which is the same for any point on the ring.

That allows you to do this calculation without using integration:
$$dE_z = |d\vec E|\cos \alpha = \frac{1}{4\pi \epsilon_0}\frac{dq}{r^2}\cos \alpha$$
And, as $z$ is fixed, everything is constant, so:
$$E_z = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\cos \alpha$$
And you can express $Q, r$ and $\alpha$ in terms of $\lambda, R$ and $z$ to get your answer.

Of course, you can integrate round ther ring if you want to, but it must give simply the total charge $Q = 2\pi R\lambda$.

But they make me give it up to $z$, don't they?

 

[PeroK]

But they make me give it up to $z$, don't they?

I'm not sure what that means.

 

[vela]

In $x$ axis the field is $0$. We calculate the electric field:
$$dE=k\dfrac{dq}{r^2}\rightarrow dq=\lambda dl$$

To be more precise, this is the magnitude of the electric field due to the infinitesimal charge $dq$.

$$E=k\int_0^L \dfrac{\lambda dl}{r^2}=\dfrac{k\lambda}{r^2}\int_0^L dl=\dfrac{k\lambda}{r^2}2\pi R$$

This is incorrect because you generally don't add vectors by adding magnitudes. You can, however, sum the components.

Question: I have problems with these exercises with integrals. For example, I understand this, but the drawing of the principle I don't see because it takes two instead of one

I don't know what you mean "it takes two instead of one," but I'll guess you're referring to the two vectors in the drawing. I think if you recognize that $\vec E$ is a vector field, it may clear up your confusion.

and in the integral I don't understand because in some exercises there are things that are constant and things that are not.

Can you give an example? It's usually pretty straightforward to decide if a quantity in the integral is a constant or not.

 

[Guillem_dlc]

Can you give an example? It's usually pretty straightforward to decide if a quantity in the integral is a constant or not.

For example in this it made me make a different integral you know

 

[kuruman]

For example in this it made me make a different integral you know

The integral is really no different in the two cases. The general procedure is to write a vector element of the field use the general form $$d\vec E=\frac{kdq(\vec r -\vec {r}')}{|\vec r -\vec {r}'|^3}$$where $\vec r$ is the position vector of the point where the electric field is to be found and $\vec {r}'$ is the position vector of the element of charge $dq$. Integrals are to be done over the primed variables.
First, you set up the position vectors. In the ring example
$\vec r = z~\hat k~;~~\vec {r}'=R\cos\phi '~\hat i+R\sin \phi '~\hat j~.~~$ Then
$(\vec r -\vec {r}')=-R\cos\phi '~\hat i-R\sin \phi '~j +z~\hat k~~$ and
$|\vec r -\vec {r}'|^3=\left[R^2+z^2\right]^{3/2}~~$so that $$d\vec E=\frac{kdq(-R\cos\phi '~\hat i-R\sin \phi '~j +z~\hat k)}{\left[R^2+z^2\right]^{3/2}}.$$Second, you find the components of the field formally by integration over the primed coordinates, in this case $\phi '$ after setting $dq=\frac{Q}{2\pi}d\phi '$
$$\begin{align} & E_x=\frac{kQ}{2\pi}\int_0^{2\pi} \frac{(-R\cos\phi ')d\phi '}{\left[R^2+z^2\right]^{3/2}}=0 \nonumber \\ & E_y=\frac{kQ}{2\pi}\int_0^{2\pi} \frac{(-R\sin\phi ')d\phi '}{\left[R^2+z^2\right]^{3/2}}=0 \nonumber \\ & E_z=\frac{kQ}{2\pi}\int_0^{2\pi} \frac{z d\phi '}{\left[R^2+z^2\right]^{3/2}}= \frac{ kQz }{\left[R^2+z^2\right]^{3/2}}\nonumber \end{align}.$$The linear charge distribution follows the same method except in one dimension.