how to solve this electric field question?

Question

a 10cm long thin rod uniformly charged to +10nC and a 10cm long thin rod uniformly charged to -10nC are 4cm apart. what are electric field strengths E1 to E3 at distances 1cm 2cm and 3cm from the glass rod along the line connecting the midpoints of the 2 rods? i know the formula for an infinite line of charge is E=1/(4*pi*epsilon 0) * 2L/r where epsilon 0=8.85e-12 or E=2KL/r where K=9.0e9.

sorry in the equation L should be lamda (linear charge density)

Answer 1

E=1/(4*pi*eo) * 2 λ/r is for infinite rod (L >>> y)
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for bisector point (distance y from +rod and (0.04 - y) from negative rod) fields are

E(y) = 9*10^9[2 λ/y] [(L/2) /sqrt{y^2+(L/2)^2}] direction (+x)

E(0.04-y)=9*10^9[ -2λ/(0.04-y)][(L/2) /sqrt{(0.04-y)^2+(L/2)^2}]
note )- ve sign of lamda λ in 2nd equ >> indicates opposite direction of field due to -ve rod

Net field at point (y)
E(y-net) = E(y) (i) + E(0.04-y) ( - i)
|E(y-net)} = |E(y)| - |E(0.04-y) --- (A)
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λ = Q/L = 10*10^-9/0.1 = C/meter
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for y=1cm = 0.01 meter
calculate E(0.01) then (0.04 - 0.01) by 2 equs and use (A)
repeat for y=2 y=3

Answer 2

Take a look at one rod. Put the origin at its center, then move right, call that x. Moving up from the center call that y, so that y increases with the length of the rod.

The rod has a linear charge density of L = 10nC/cm. A small chunk of charge along the length of the rod, dq, is:

dq = Ldy

The field magnitude corresponding to a point along x due to this charge is:

dE = (1/(4pi*eps)) * (dq/(y^2+x^2))

dE = (1/(4pi*eps)) * (Ldy/(y^2+x^2))

This is the magnitude, and we want the direction along x, so multiply by the cos of the angle that this field chunk makes with the x axis:

cos(theta) = x/(x^2+y^2)^(1/2)

This makes the x component of the field:

dE(x) = (1/(4pi*eps)) * (Lxdy/(y^2+x^2)^(3/2))

To check our result, we see that in the limit of x-->infinity, when the rod looks more like a point, that the field falls off as x^(-2) as it should.

There is no y component due to symmetry.

To find the total force, you could integrate dE(x) from y =-10cm . . 10cm, or from 0 to 10 and multiply by two. This will give you E(x) for a rod.