The direction (theta) of the electric field v | vedclass

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(A) For a uniformly charged finite rod,the electric field components at a point $P$ are given by $E_x = \frac{k\lambda}{d}(\sin 	heta_1 + \sin 	heta

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at an angle of $30^{\circ}$ from the $x$-axis

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(A) For a uniformly charged finite rod,the electric field components at a point $P$ are given by $E_x = \frac{k\lambda}{d}(\sin 	heta_1 + \sin 	heta_2)$ and $E_y = \frac{k\lambda}{d}(\cos 	heta_2 - \cos 	heta_1)$.In the given figure,the rod subtends angles $	heta_1 = 90^{\circ}$ and $	heta_2 = 30^{\circ}$ with the perpendicular line from $P$ to the rod.Thus,$E_x = \frac{k\lambda}{d}(\sin 90^{\circ} + \sin 30^{\circ}) = \frac{k\lambda}{d}(1 + 0.5) = 1.5 \frac{k\lambda}{d}$.And $E_y = \frac{k\lambda}{d}(\cos 30^{\circ} - \cos 90^{\circ}) = \frac{k\lambda}{d}(\frac{\sqrt{3}}{2} - 0) = 0.866 \frac{k\lambda}{d}$.The angle $	heta$ of the resultant electric field with the $x$-axis is given by $	an 	heta = \frac{E_y}{E_x} = \frac{0.866}{1.5} = \frac{\sqrt{3}/2}{3/2} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Therefore,$	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.

The direction $(\theta)$ of the electric field $\vec{E}$ at point $P$ due to a uniformly charged finite rod is shown in the figure. Based on the geometry provided,determine the angle $\theta$.

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