91Ó°ÊÓ

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

Required:

a) E, localid="1665051380337" $\mathrm{θ}$

b) F, $\mathrm{θ}$

a) To not perform many calculations let's compare the equation $\left(\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{o}x\sqrt{{\mathrm{Î\pm }}^2+x^2}}\hat{i}\right)$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

$E_{+}=\frac{1}{x\sqrt{{\mathrm{Î\pm }}^2+x^2}}$

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

$E_{-}=\frac{1}{y\sqrt{{\mathrm{Î\pm }}^2+y^2}}$

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

Final expression: The final expression for electric field is

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

Substitution in (1) gives,

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

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

The Magnitude of E is

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

The angle is given by

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

But $\mathrm{θ}$ is in 3^rd quadratic based on our assumption

$\mathrm{θ}=45+180=225°$

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

$F=eE=1.6×{10}^{-19}×6.2×{10}^4≈1.0×{10}^{-14}$

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°.