Charge on 1cm Wire: Solve with E=Q/4πεor^2

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SUMMARY

The discussion focuses on calculating the charge on a 1.0-cm-long segment of a charged wire given the electric field at a distance of 4.80 cm. The correct approach involves using the formula E = (2λ)/(4πε₀r), where λ represents the linear charge density. The user initially attempted to use an incorrect formula but was guided to derive the electric field in terms of charge density. This method leads to the correct calculation of charge using the relationship λ = Q/L.

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Homework Statement


The electric field 4.80 cm from a very long charged wire is ( 2100 N/C,toward the wire).
What is the charge (in nC) on a 1.0-cm-long segment of the wire?

Homework Equations



E=1/(4*pi*eo) * Q/ri^2 * cos (theta)

3. The Attempt at a Solution

I put in 9*10^9 for 1/(4*pi*eo)
I put in (.048^2+.01^2) for ri^2 and for cos theta i put .01/4.8

I thought this equation would have worked with the old plug and chug method, is this the right equation for this situation?

Thanks
 
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No, that's not correct. Think about what the equation means and you'll see why.

To start off, assume that the line of charge has a charge density of lambda and use that to find an equation giving the electric field in terms of r. Then plug and chug into THAT equation.
 
Ah. 2(lambda)/(4*pi*eo*r)
lambda = Q/L
Got it.
Thanks!
 

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