Charge on Semicircle: Electric Field Equation

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In summary, the conversation discusses the difference in calculating the electric field for a charge on a semicircle compared to a ring. The vertical component can still be calculated easily, but the other component will require its own integral. The conversation also touches on the direction of the electric field and the need for showing the full work to find any errors.
  • #1
freezer
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Homework Statement


This is not a homework problem so there is no problem statement. More of a conceptual question.

Consider a charge on a ring:
eler3.gif


If this was a semicircle opposed to a ring, how would this change the equation since there is not an opposing dEr force.

[tex]

\boldsymbol{E} = \boldsymbol{\hat{z}} \frac{\rho_l R(-\hat{\boldsymbol{r}}R + \hat{\boldsymbol{z}Z})}{4 \pi \varepsilon_0 (R^2 + Z^2)^{3/2}}\int_{0}^{\pi }d\phi

[/tex]
 
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  • #2
The vertical component is still easy to calculate (as you just have a smaller part of the ring contributing to it), but the other component (due to the broken symmetry it does not vanish any more) will need its own integral. No idea how easy/messy that gets.
 
  • #3
So I would have to do it twice.

dE = dEr + dEz
 
  • #4
The integrals are different, but yes.
 
  • #5
Does this look reasonable:

[tex]

\frac{Q}{4\pi \varepsilon_0(R^2 + Z^2)^{3/2} }(Z\hat{z}-R\hat{r})

[/tex]
 
  • #6
Z and z are the same?

That would mean the electric field points in the same direction as the "r" line in the sketch. No, that cannot work.
It would be easier to find the error if you show your whole work.
 

1. What is the equation for the electric field on the semicircle?

The equation for the electric field on the semicircle is given by E = kQ/R, where k is the Coulomb's constant, Q is the charge on the semicircle, and R is the radius of the semicircle.

2. How does the charge on the semicircle affect the electric field?

The electric field on the semicircle is directly proportional to the charge on the semicircle. This means that as the charge increases, the electric field also increases.

3. What is the direction of the electric field on the semicircle?

The electric field on the semicircle is always perpendicular to the tangent of the semicircle at any given point. This means that the direction of the electric field changes as you move along the semicircle.

4. Can the electric field on the semicircle be negative?

Yes, the electric field on the semicircle can be negative. This can occur when the charge on the semicircle is negative, resulting in an electric field that points in the opposite direction of the electric field due to a positive charge.

5. How does the radius of the semicircle affect the electric field?

The electric field on the semicircle is inversely proportional to the radius of the semicircle. This means that as the radius increases, the electric field decreases. This is because as the distance between the charges on the semicircle increases, the electric field weakens.

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