How to Calculate the Electric Field at the Origin from a Uniformly Charged Rod?

[J6204]

Homework Statement

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

Homework Equations

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

 

[TSny]

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

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

 

[J6204]

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

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?

 

[TSny]

I thought so, do the answers look correct besides one or two of them being negative?

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.