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(A) The $x-$component of the electric field due to a finite rod at a point $P$ is given by $E_x = \frac{k\lambda}{r}(\sin 	heta_2 - \sin 	heta_1)$.H

Vedclass · Accepted Answer

$3.6	imes10^7\, N/C$

Vedclass · Answer

(A) The $x-$component of the electric field due to a finite rod at a point $P$ is given by $E_x = \frac{k\lambda}{r}(\sin 	heta_2 - \sin 	heta_1)$.Here,$r = 3\,cm = 0.03\,m$,$\lambda = 30\,\mu C/m = 30 	imes 10^{-6}\,C/m$,and $k = 9 	imes 10^9\,N\cdot m^2/C^2$.The angles are $	heta_1 = 0^{\circ}$ (at the end directly below $P$) and $	heta_2 = 53^{\circ}$ (at the other end).Substituting the values:$E_x = \frac{(9 	imes 10^9)(30 	imes 10^{-6})}{0.03} (\sin 53^{\circ} - \sin 0^{\circ})$$E_x = \frac{27 	imes 10^4}{0.03} (0.8 - 0)$$E_x = 9 	imes 10^6 	imes 0.8 = 7.2 	imes 10^6\,N/C$.Wait,re-evaluating the geometry: The formula for the $x-$component is $E_x = \frac{k\lambda}{r}(\cos 	heta_1 - \cos 	heta_2)$ where $	heta$ is measured from the perpendicular. With $	heta_1 = 0$ and $	an 	heta_2 = 4/3$,$	heta_2 = 53^{\circ}$.$E_x = \frac{9 	imes 10^9 	imes 30 	imes 10^{-6}}{0.03} (\cos 0^{\circ} - \cos 53^{\circ})$$E_x = 9 	imes 10^6 (1 - 0.6) = 9 	imes 10^6 	imes 0.4 = 3.6 	imes 10^6\,N/C$.

$A$ uniformly charged rod of length $4\,cm$ and linear charge density $\lambda = 30\,\mu C/m$ is placed as shown in the figure. Calculate the $x-$ component of the electric field at point $P$.

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