- #1
al_famky
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calculating electric fields due to continuous charge distributions?
a question I came across doing some electric field questions, and the answer was really confusing.
Charge is distributed along a linear semicircular rod with a linear charge density λ as in picture attatched. Calculate the electric field at the origin
for the first question, the magnitude of E was calculated with dE=[itex]\frac{adθλ}{4\pi\epsilon_{0}a}[/itex], but why is there only one "a" in the denominator, not [itex]a^{2}[/itex]? The final result was this equation [itex]\int[/itex][itex]^{\phi}_{-\phi}[/itex][itex]\frac{λcosθdθ}{4\pi\epsilon_{0}}[/itex]=[itex]\frac{λsin\phi}{2\pi\epsilon_{0}}[/itex] which makes sense only if the "a" was dropped.
Thanks to anyone who'd be willing to explain the answers!
a question I came across doing some electric field questions, and the answer was really confusing.
Homework Statement
Charge is distributed along a linear semicircular rod with a linear charge density λ as in picture attatched. Calculate the electric field at the origin
Homework Equations
The Attempt at a Solution
for the first question, the magnitude of E was calculated with dE=[itex]\frac{adθλ}{4\pi\epsilon_{0}a}[/itex], but why is there only one "a" in the denominator, not [itex]a^{2}[/itex]? The final result was this equation [itex]\int[/itex][itex]^{\phi}_{-\phi}[/itex][itex]\frac{λcosθdθ}{4\pi\epsilon_{0}}[/itex]=[itex]\frac{λsin\phi}{2\pi\epsilon_{0}}[/itex] which makes sense only if the "a" was dropped.
Thanks to anyone who'd be willing to explain the answers!
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