Find the electric field

+1 vote

A line of charge with uniform density of 30.0 nC/m lies along the line y = -15.0 cm, between the points with coordinates x = 0 and x = 32.0 cm. Find the electric field it creates at the origin. 

 
asked Sep 4, 2012 in Physics by Kiera12 (46 points)
    

1 Answer

+1 vote

E_x = 1/(4πε₀) ∫ λ dx * 1/ (Y² + x²) * x/((Y² + x²)^½

where Y is the vertical distance 15 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)" ).

The integral is easy and gives

E_x = λ/(8πε₀) ∫ du * 1/ u^(3/2)
= λ/(8πε₀) [-2/√u]
= λ/(4πε₀) (1/Y - 1/√(Y² + X²)) [where X = 32.0 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))


Just substitute λ (30*10^-9 C/m), X ( 0.30 m ) and Y ( 0.15 m) and ε₀ ( 8.854 10^-12 F/m) to calculate an uninteresting set of two numbers (E_x and E_y)

answered Nov 3, 2012 by talat ameen ~Rookie~ (80 points)



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