A point charge of 10,mu C is placed at the or | vedclass

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(A) Let the charge $q_1 = 10\,\mu C$ be at $x_1 = 0$ and the charge $q_2 = 40\,\mu C$ be at $x_0$.The electric field at point $P$ $(x = 2\,cm)$ due to

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$x = 6\,cm$

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(A) Let the charge $q_1 = 10\,\mu C$ be at $x_1 = 0$ and the charge $q_2 = 40\,\mu C$ be at $x_0$.The electric field at point $P$ $(x = 2\,cm)$ due to $q_1$ is $E_1 = \frac{K q_1}{r_1^2} = \frac{K 	imes 10}{(2)^2}$ (directed towards the right).The electric field at point $P$ due to $q_2$ is $E_2 = \frac{K q_2}{r_2^2} = \frac{K 	imes 40}{(x_0 - 2)^2}$ (directed towards the left).For the net electric field at $P$ to be zero,$E_1 = E_2$.$\frac{K 	imes 10}{2^2} = \frac{K 	imes 40}{(x_0 - 2)^2}$$\frac{10}{4} = \frac{40}{(x_0 - 2)^2}$$(x_0 - 2)^2 = \frac{40 	imes 4}{10} = 16$Taking the square root on both sides,$x_0 - 2 = 4$ (since the charge must be to the right of $P$ for the fields to cancel).$x_0 = 6\,cm$.

$A$ point charge of $10\,\mu C$ is placed at the origin. At what location on the $X$-axis should a point charge of $40\,\mu C$ be placed so that the net electric field is zero at $x = 2\,cm$ on the $X$-axis?

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