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
Ey = integ { cos(q) dq / [ 4 p e0 R2] }
dq = l dx
If your computer doesn't have the "symbol" font, the equations read:
Ey = 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
Ey =
integ{k l y dx /
r3 }
Ey =
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
Ey = k l y integ[ (x2 + y2 ) - 3/2 dx] Ey = 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
Ey = k l y integ[1/sec(q)] Ey = k lambda y integ[1/sec(theta)], or
Ey = k l y integ[cos(q)] Ey = k lambda y integ[cos(theta)]
Ey = k l y sin(q) Ey = 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
Ey = k l a / [y sqrt (y2 = a2) ] Ey = k lambda a / [y sqrt (y2 = a2) ]