Two particles with charges +Q and -Q are plac | vedclass

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(C) Let the charges be at points $A$ and $C$ with distance $r$. The force between them is $F = k\frac{Q^2}{r^2}$.When a charge $+Q$ is placed at point

Vedclass · Accepted Answer

$8F$, in the direction of $-Q$ charge

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(C) Let the charges be at points $A$ and $C$ with distance $r$. The force between them is $F = k\frac{Q^2}{r^2}$.When a charge $+Q$ is placed at point $B$ (midpoint), the distance from $A$ to $B$ is $r/2$ and from $B$ to $C$ is $r/2$.The force exerted by charge at $A$ on $B$ is $F_A = k\frac{Q \cdot Q}{(r/2)^2} = 4k\frac{Q^2}{r^2} = 4F$ (repulsive, towards $C$).The force exerted by charge at $C$ on $B$ is $F_C = k\frac{|Q \cdot (-Q)|}{(r/2)^2} = 4k\frac{Q^2}{r^2} = 4F$ (attractive, towards $C$).Since both forces are in the same direction (towards the $-Q$ charge), the net force is $F_{net} = F_A + F_C = 4F + 4F = 8F$ in the direction of the $-Q$ charge.

Two particles with charges $+Q$ and $-Q$ are placed at a certain distance. The force between them is $F$. If a particle with charge $+Q$ is placed exactly in the middle of these two particles, what is the net force acting on it?

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