Two particles A and B having charges 20, mu C | vedclass

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(B) Let the third charge $q$ be placed at a distance $x$ from the charge $-5\, \mu C$ on the side away from the $20\, \mu C$ charge.For the net electr

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At $5\, cm$ from $-5\, \mu C$ on the right side

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(B) Let the third charge $q$ be placed at a distance $x$ from the charge $-5\, \mu C$ on the side away from the $20\, \mu C$ charge.For the net electric force to be zero,the electric field at that point must be zero.The electric field due to the $20\, \mu C$ charge at distance $(5+x)$ and the $-5\, \mu C$ charge at distance $x$ must be equal in magnitude and opposite in direction.$\frac{k(20\, \mu C)}{(5+x)^2} = \frac{k(5\, \mu C)}{x^2}$$\frac{20}{(5+x)^2} = \frac{5}{x^2}$Taking the square root on both sides:$\frac{\sqrt{20}}{5+x} = \frac{\sqrt{5}}{x}$$\frac{2\sqrt{5}}{5+x} = \frac{\sqrt{5}}{x}$$2x = 5+x$$x = 5\, cm$Thus,the point is at $5\, cm$ from the $-5\, \mu C$ charge on the right side.

Two particles $A$ and $B$ having charges $20\, \mu C$ and $-5\, \mu C$ respectively are held fixed with a separation of $5\, cm$. At what position should a third charged particle be placed so that it does not experience a net electric force?

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