Question

An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis at y = 2.0 m. The x component of the electric field at the point P on the x axis at x = 3.0 m is approximately
A) 1.8 kN/C B) 4.2 kN/C C) 0.96 kN/C D) 5.2 kN/c E) 0.64 mN/C

Answer 1

i calculate the field from the point charge q:

\[E = \frac{1}{4.\pi.\epsilon _{0}}.\frac {q}{r ^{2}}.r ^{.}\]
epsilon is permittivity of free space, r distance to point charge, q is charge, and r point is the unit vector to the point p (q1->P).
so
\[E = \frac{1}{4.\pi.8,85.10E-12}.\frac {6µC}{3,6^{2}}.r ^{.}\]
is
E= 4162,87 N/C = 4,16kN/C

Logically the E field must be still bigger than this number, since there is still the charges on the Z-axis, (also positive charges).
So the answer must be either B or D.
Now I must integrate the charges on Z-axis for to obtain the field in P, than add to the previous result.

Answer 2

\[E = \frac {1}{4.\pi.\epsilon _{0}} . 2. \int\limits_{0}^{\infty} \frac {dq}{r²} = \frac {1}{4.\pi.\epsilon _{0}} .2.\int\limits_{0}^{\infty} \frac {0,3µC/m.dz}{z²+3²}\]
which means i have to integrate from z= 0 to z= infinity, since the design is symmetric over the z axis, I multiply by 2 (same result as integrating from -infinity to + infinity), where dq is the charge from an infinitely small point dz on the z axis.
also distance of r² = 3²+z² (pythagoras right triangle)