Electric field due to arc shaped thin rod

[microdosemishief]

Homework Statement

Please see image

Relevant Equations

Coulomb's Law (dE=kdQ/r^2)

My attempt: due to symmetry along x-axis, net E is only along x^hat.

dQ = λ dl = λ (R dθ)
for each dl, the x component of distance from dl to the origin is Rcos(θ)

Hence, E_x = \int_{-θ_0}^{θ_0} k(λ R dθ)/(Rcos(θ)^2 = kλ/R \int_{-θ_0}^{θ_0} sec(θ)^2 dθ = 2kλ/R tan(θ_0) along negative x^hat


But the correct answer has sin(θ_0) instead of tan(θ_0)

 

[TSny]

For the element of charge $dQ$ shown, how would you write an expression for $dE$ at the origin?
How would you write the x-component of $dE$?

 

[microdosemishief]

I wrote that in my question

 

[TSny]

I wrote that in my question

What was your reason for writing the denominator of the integrand as $(R\cos \theta)^2$? This is not correct.

To find $E_x$ due to the entire charge $Q$, you need to understand that $E_x = \int dE_x$ where $dE_x$ is the x-component of the field vector $\overrightarrow{dE}$ due to an arbitrary element of charge $dQ$.

So, first, think about the vector $\overrightarrow{dE}$ at the origin produced by $dQ$. You should draw the vector $\overrightarrow{dE}$ at the origin in a diagram and write an expression for the magnitude, $dE$, of this vector.

Next, think about how you can write an expression for the x-component, $dE_x$, of the vector $\overrightarrow{dE}$. Then you will be ready to set up the integration $E_x = \int dE_x$.

 

[microdosemishief]

Yes and dE = kdQ/r^2 so the denominator is r^2 where r = Rcos(θ) squared, because dE_x is Rcosθ distance away

 

[TSny]

dE = kdQ/r^2 so the denominator is r^2 where r = Rcos(θ) squared, because dE_x is Rcosθ distance away

No. Let's forget dE_x for now. Just concentrate on dE.

dE is the magnitude of the electric field vector dE at the origin O produced by the charge dQ in the diagram in post #2. In the general formula dE = kdQ/r², what would you use for the distance r in this case?

 

[microdosemishief]

R

 

[microdosemishief]

I just read the textbook answer but I want to know why my answer is wrong and not just a different method to the same answer

 

[TSny]

R

yes

 

[TSny]

I just read the textbook answer but I want to know why my answer is wrong and not just a different method to the same answer

Your answer is wrong because you can't get the x-component dE_x by replacing R in the denominator of dE by Rcosθ.

Note that Rcosθ < R when θ≠0. So, kdQ/(Rcosθ)² is greater than kdQ/R² when θ≠0. That is, your expression for dE_x is greater than the expression for dE. But the component of a vector cannot be greater than the magnitude of the vector.

 

[TSny]

Yes and dE = kdQ/r^2 so the denominator is r^2 where r = Rcos(θ) squared, because dE_x is Rcosθ distance away

I don't understand why you say that "dE_x is Rcosθ distance away". Both dE and dE_x are associated with the same point, namely the origin:

In what sense is dE_x located a distance Rcosθ away?

 

[kuruman]

I just read the textbook answer but I want to know why my answer is wrong and not just a different method to the same answer

Your answer $$E=\frac{2k\lambda}{R}\tan\theta_{0}$$ is wrong because for a quarter circle $\left(\theta_0=90^{\circ}\right)$ it implies that the electric field at the center will be infinite which doesn't make sense. The field is finite in the range $0<\theta_0<2\pi.$

 

[weirdoguy]

$Rcos\theta$ is x component of vector $\vec{r}$ not $d\vec{E}$, where by $\vec{r}$ i mean vector that starts at the origin, and points do $dQ$.

 

[kuruman]

$Rcos\theta$ is x component of vector $\vec{r}$ not $d\vec{E}$, where by $\vec{r}$ i mean vector that starts at the origin, and points do $dQ$.

The electric field from a charge element $dq=\lambda d\theta$ is a vector $d\vec E$ which has magnitude $$dE=\frac{kdq}{R^2}=\frac{k\lambda d\theta}{R^2}$$ because the distance between the element and the point of interest is $R$. All elements $dq$ are at the same distance $R$ from the center of the arc. The magnitudes of their contributions $d\vec E$ are the same but their directions are different. To find the net field you need to add all elements $d\vec E$ as vectors, which means that you add all the x-components together and, separately, all the y-components together. That's called superposition.

So what are $dE_x$ and $dE_y$ if the magnitude is $dE$ as given above?

On edit
If you insist on using vector $\mathbf r$, you can start formally from the general expression for the electric field at position $\mathbf r$ due to a point charge $Q$ at position $\mathbf r'$, $$\mathbf E=\frac{kQ(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r '|^3}.$$ Then substitute
$\mathbf r=0$; $~\mathbf r'=R\cos\theta~\mathbf{\hat x}+R\sin\theta~\mathbf{\hat y}$; $~Q=\lambda d\theta$ and integrate. You will get the same answer.