- #1
dimensionless
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- 1
I have a circular arc of wire centered at the point (0,0). It has a radius of [tex]r[/tex], extends from [tex]\theta = -60[/tex] to [tex]\theta = 60[/tex] and also holds a charge [tex]q[/tex]. For the differential electric field I have the following equation:
[tex]
dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}
[/tex]
Where [tex]ds[/tex] is the length of a differential element of the arc.
To the find the [tex]x[/tex] component of the electric field I this equation:
[tex]
dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds
[/tex]
To integrate this I have to set [tex]ds = r d\theta[/tex] so that the above equation reads:
[tex]
dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta
[/tex]
Where does the [tex]r[/tex] come from in the statement [tex]ds = r d\theta[/tex]?
[tex]
dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}
[/tex]
Where [tex]ds[/tex] is the length of a differential element of the arc.
To the find the [tex]x[/tex] component of the electric field I this equation:
[tex]
dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds
[/tex]
To integrate this I have to set [tex]ds = r d\theta[/tex] so that the above equation reads:
[tex]
dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta
[/tex]
Where does the [tex]r[/tex] come from in the statement [tex]ds = r d\theta[/tex]?