# E field of a long charged rod

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1. Jun 21, 2017

### J6204

1. The problem statement, all variables and given/known data

A line of charge with a uniform density of 34.2 nC/m lies along the line y = -14.9 cm, between the points with coordinates x = 0 and x = 42.8 cm. Calculate the electric field it creates at the origin, entering first the x component then the y component
2. Relevant equations

3. The attempt at a solution
E_x = 1/(4πε₀) ∫ λ dx * 1/ (Y² + x²) * x/((Y² + x²)^½

where Y is the vertical distance 14.9 cm.
λ dx is the charge element dq, 1/ (Y² + x²) is the " 1/r^2 " and x/((Y² + x²)^½ is the geometric factor for the x-component ("sin(α)" ).

Then the integral gives

E_x = λ/(8πε₀) ∫ du * 1/ u^(3/2)
= λ/(8πε₀) [-2/√u]
= λ/(4πε₀) (1/Y - 1/√(Y² + X²)) [where X = 42.8 cm]

Along the same line of reasoning we have for the y-component

E_y = 1/(4πε₀) ∫ λ dx * 1/ (Y² + x²) * Y/((Y² + x²)^½
= Yλ/(4πε₀) ∫ dx /(Y² + x²)^(3/2)
= λ/(4πε₀) X/(Y√(Y² + X²))

When i substituted λ (34.2*10^-9 C/m), X ( 0.428 m ) and Y ( 0.149 m) and ε₀ ( 8.854 10^-12 F/m) to calculate Ex and Ey I got the following two numbers which were incorrect and I am not sure what I am doing wrong

Ex = 1385N/C
Ey = 1948N/C

2. Jun 21, 2017

### TSny

Your approach looks good. I have not checked the numerical evaluation. But do you expect both components of E to be positive?

3. Jun 21, 2017

### J6204

I thought so, do the answers look correct besides one or two of them being negative? Did I need to factor in the negative sign on the y coordinate?

4. Jun 21, 2017

### TSny

Yes. To determine the signs of the components, choose an arbitrary point along the line of charge and treat the point as a positive point charge. Consider the direction of E at the origin produced by the point charge.

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