Assignment 4                                                 Solutions

10. List of quantities in the problem

Linear charge density	lambda = const.
Length of rod	a
y component of electric field	Ey

Physics

Integrate over Coulomb's Law

Working equations

E_y = integ { cos(q) dq / [ 4 p e₀ R²] }

dq = l dx

If your computer doesn't have the "symbol" font, the equations read:

E_y = integ { cos(theta) dq / [ 4 pi epsilon₀ R²] }

dq = lambda dx

For the rest of the problem, I will write
k = 1 / (4 pi epsilon₀ )

Details

Substituting for dq, and using

cos(q) = y / r        cos(theta) = y / r

gives

E_y = integ{k l y dx / r³ }        E_y = integ{k lambda y dx / r³ }
with the integral running from x=0 to x=a

From the diagram, r = sqrt[x² + y²], which makes

E_y = k l y integ[ (x² + y² ) ^{- 3/2} dx]        E_y = k lambda y integ[ (x² + y² ) ^{- 3/2} dx]

The integral is accomplished by the substitution

x = y tan (q)        x = y tan (theta)

dx = y sec² (q)        dx = y sec² (theta)

which gives

E_y = k l y integ[1/sec(q)]        E_y = k lambda y integ[1/sec(theta)], or

E_y = k l y integ[cos(q)]        E_y = k lambda y integ[cos(theta)]

E_y = k l y sin(q)        E_y = k lambda y sin(theta)

where the expression must be evaluated at x=a and at x=0. At those two locations, the diagram shows that the sin is a / sqrt(y² + a² and 0, respectively, so that

E_y = k l a / [y sqrt (y² = a²) ]        E_y = k lambda a / [y sqrt (y² = a²) ]