A very thin line of charge lies along the z axis from x=-infty to x=+infty Another similar line of charge lies along the y axis from y=-infty to y=+infty Both lines have a uniform charge per length lambda Part A (lambda )/(2pi epsilon _(0)xy)sqrt (x^2+y^2) Determine the resulting electric field magnitude (relative to the z axis) at a point (x,y) in the first quadrant of the zy plane (lambda )/(2pi varepsilon _(0)x^2)sqrt ((1)/(x^2)+y^2) (lambda )/(2varepsilon _(0)y^2)sqrt (x^2+y^2) (lambda )/(pi varepsilon _(0)xy^2)sqrt (x^2+y^2)

The correct answer is:<br /><br />$\frac {\lambda }{2\pi \varepsilon _{0}xy}\sqrt {x^{2}+y^{2}}$<br /><br />Explanation:<br />The electric field due to a line of charge along the z-axis at a point $(x,y)$ is given by:<br /><br />$E_z = \frac {\lambda }{2\pi \varepsilon _{0}xy}\sqrt {x^{2}+y^{2}}$<br /><br />The electric field due to a line of charge along the y-axis at a point $(x,y)$ is given by:<br /><br />$E_y = \frac {\lambda }{2\pi \varepsilon _{0}xy}\sqrt {x^{2}+y^{2}}$<br /><br />Since both lines have a uniform charge per length $\lambda$, the resulting electric field magnitude (relative to the z axis) at a point $(x,y)$ in the first quadrant of the zy plane is the sum of the magnitudes of $E_z$ and $E_y$, which is:<br /><br />$E = \sqrt{E_z^2 + E_y^2} = \frac {\lambda }{2\pi \varepsilon _{0}xy}\sqrt {x^{2}+y^{2}}$<br /><br />Therefore, the correct answer is $\frac {\lambda }{2\pi \varepsilon _{0}xy}\sqrt {x^{2}+y^{2}}$.