What is the Electric Field at a Point Near an Infinite Line of Charge?

AI Thread Summary
The discussion focuses on calculating the electric field at the point (0, -3 m) due to an infinite line of negative charge with a uniform charge density of λ = -2.3 μC/m. The x-component of the electric field is determined to be zero since the charge lies along the y-axis. Participants suggest using the formula E = 2kλ/r, but clarify that it is not suitable for this scenario as it pertains to points along the line of charge. Instead, they recommend setting up an integral for the electric field contribution from an element of charge, with the integral limits from 0 to infinity. The correct integrand is provided as ∫(0 to ∞) kλ/(y+3)^2 dy, emphasizing the importance of focusing on the magnitude of the charge.
TwinGemini14
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An infinite line of negative charge begins at the origin and continues forever in the +y-direction. It has a uniform charge distribution of λ = -2.3 μC/m. Calculate the x and y-component of the electric field at the point (0,-3 m).

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So immediately realized that Ex = 0 since te charge also lies on the y axis. I cannot seem to evaluate what Ey would be.

L=Lambda

Here's what I'm doing: E = 2kL / r
E = 2(9*10^9)(-2.3*10^-6) / (-3)
E = 13800 N/C

What am I doing wrong? Can somebody please help me? I've been working on this problem for a very long time. Thanks in advance.
 
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TwinGemini14 said:
Here's what I'm doing: E = 2kL / r
That formula describes the field surrounding an infinite line of charge at some distance (r) from the line. Not useful for this problem since you are finding the field along the line of the charge and you are dealing with the end of the line of charge. Instead, set up an expression for the field contributed by an element of charge and integrate.
 
I don't understand how to set up this integral. Will the limits be from 0 to infinity?

inf.
k | (-2.3*10^-6)y / (3+y)^2
0


| = integrand

Does this look right? I seriously do not understand how to set up this equation. Help please!
 
TwinGemini14 said:
inf.
k | (-2.3*10^-6)y / (3+y)^2
0
That's pretty close. Here's how I would write it:

(0 to ∞) kλ/(y+3)^2 dy

Forget the sign of the charge, just worry about the magnitude. You'll assign the proper direction and sign to the field at the end.
 
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