Two 1.20 m non-conducting rods meet at a right angle. One rod carries +2.50 mC of charge distributed uniformly along its length, and the other carries -2.50 mC distributed uniformly along with it (in Fig.).

(a) Find the magnitude and direction of the electric field these rods produce at point P, which is 60.0 cm from each rod.

(b) If an electron is released at P, what are the magnitude and direction of the net force that these rods exert on it?

$\mathrm{Q}=2.5\times {10}^{-6}\mathrm{C}$, a=0.6 m, $\mathrm{x}$=0.6 m, $\mathrm{y}$=0.6 m

Required:

a) E, localid="1665051380337" $\theta$

b) F, $\theta$

a) To not perform many calculations let's compare the equation $\left(\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_{o}x\sqrt{{\alpha }^2+x^2}}\hat{i}\right)$using superposition to get the final expression of the electric field at point p

For horizontal rod: The rod is on the y-axis and locates from -a to +a, x-aris is perpendicular on it with positive direction to upward, so the rod will have only one component in – x-direction so the magnitude of E due to +ve distribution is

$E_{+}=\frac{1}{x\sqrt{{\alpha }^2+x^2}}$

And the direction is $\hat{i}$.

For vertical rod: The rod is on the x-axis and locates from - a to +a, the y-axis is perpendicular to it with a positive direction to the right, so the rod will have only one component in the y-direction because the charge distribution is negative, so the magnitude of E due to -ve distribution is

$E_{-}=\frac{1}{y\sqrt{{\alpha }^2+y^2}}$

And the direction is $\hat{j}$.

Final expression: The final expression for electric field is

$E=E_{+}{\hat{i}}_{+}{E}_{-}\hat{j}=\frac{1}{x\sqrt{{\alpha }^2+x^2}}\hat{i}\frac{1}{y\sqrt{{\alpha }^2+y^2}}\hat{j}$

Substitution in (1) gives,

$\mathrm{E}=-\frac{2.5\times {10}^{-6}\times 9\times {10}^9}{0.6\sqrt{{\left(0.6\right)}^2+{\left(0.6\right)}^2}}\hat{\mathrm{i}}-\frac{2.5\times {10}^{-6}\times 9\times {10}^9}{0.6\sqrt{{\left(0.6\right)}^2+{\left(0.6\right)}^2}}\hat{\mathrm{j}}$

$=\left[-4.4\times {10}^4\hat{i}-4.4\times {10}^4\hat{j}\right]N/C$

The Magnitude of E is

$E=\sqrt{E_x^2+E_y^2}=6.2\times {10}^{4}N/C$

The angle is given by

$\theta =\left(\frac{E_y}{E_x}\right)=45°$

But $\theta$ is in 3^rd quadratic based on our assumption

$\theta =45+180=225°$

b) The magnitude of the force acting on the electron is given by

$F=eE=1.6\times {10}^{-19}\times 6.2\times {10}^4\approx 1.0\times {10}^{-14}$

The electron is a negative charge so the negative distribution will push it in positive x-direction and the positive distribution will attract it in a positive y-direction so the force is in the first quadratic at an angle equal to 45°.