Two point charges +9e and +e are placed at a | vedclass

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(A) Let the charge $q$ be placed at a distance $x$ from the charge $+9e$ along the line joining the two charges.For the charge $q$ to be in equilibriu

Vedclass · Accepted Answer

$12\, cm$ from $+9e$

Vedclass · Answer

(A) Let the charge $q$ be placed at a distance $x$ from the charge $+9e$ along the line joining the two charges.For the charge $q$ to be in equilibrium,the electrostatic force exerted by $+9e$ must be equal in magnitude and opposite in direction to the force exerted by $+e$.Using Coulomb's law,$F = \frac{K q_1 q_2}{r^2}$,we set the forces equal:$\frac{K(9e)(q)}{x^2} = \frac{K(e)(q)}{(16-x)^2}$Canceling $K, e,$ and $q$ from both sides:$\frac{9}{x^2} = \frac{1}{(16-x)^2}$Taking the square root on both sides:$\frac{3}{x} = \frac{1}{16-x}$Solving for $x$:$3(16-x) = x$$48 - 3x = x$$4x = 48$$x = 12\, cm$Thus,the charge $q$ must be placed at a distance of $12\, cm$ from the $+9e$ charge.

Two point charges $+9e$ and $+e$ are placed at a distance of $16\, cm$ apart. Where should a third charge $q$ be placed between them so that it remains in equilibrium?

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