Question

Five charges are located along one straight line They are all equally spaced and the magnitudes of their charges are all the same, The three on the left are negatively charged. The fourth one is positively charged. The one farthest to the right is negatively charged. In which direction will the one in the middle, i.e. the third one, accelerate?

Answer 1

Since the charges are same in magnitude

and all charges are equally spaced,

$F_{1}=-F_{5}$

so both will cancel out

net force acting charge 3 is $F=F_{2}+F_{4}$

acting towards RIGHT

ANSWER: Towards Right

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