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(D) Let the third charge $q$ be placed at a distance $x$ from the charge $9e$. Then,its distance from the charge $16e$ is $(70 - x) \ cm$.For the thir

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Both $(a)$ and $(b)$

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(D) Let the third charge $q$ be placed at a distance $x$ from the charge $9e$. Then,its distance from the charge $16e$ is $(70 - x) \ cm$.For the third charge to be in equilibrium,the net electrostatic force on it must be zero. Therefore,the magnitude of the force exerted by $9e$ must equal the magnitude of the force exerted by $16e$:$F_1 = F_2$$\frac{k \cdot (9e) \cdot q}{x^2} = \frac{k \cdot (16e) \cdot q}{(70 - x)^2}$$\frac{9}{x^2} = \frac{16}{(70 - x)^2}$Taking the square root on both sides:$\frac{3}{x} = \frac{4}{70 - x}$$3(70 - x) = 4x$$210 - 3x = 4x$$7x = 210$$x = 30 \ cm$Thus,the charge should be placed at a distance of $30 \ cm$ from $9e$. The distance from $16e$ is $70 - 30 = 40 \ cm$. Therefore,both options $(a)$ and $(b)$ are correct.

Two electric charges are arranged as shown in the figure. Where should a third charge be placed so that it remains in equilibrium?

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