Three particles are located on the $x$-y plane as indicated in the diagram. The first particle has charge $q_{1}=e$ and is located at $<-d, 0,0>$. The second particle has charge $q_{2}=3 e$ and is located at $\langle 2 d, 0,0\rangle$ Note that $e$ is the charge of the proton and $d$ is a positive

Determine where a fourth particle with charge $q_{4}=-e$ should be positioned such that the net electric field at the origin $\langle 0,0,0\rangle$ is zero? Please answer as vector formulas using "d". 1.35 $x$ 9 $-.8$ $-.8$ $\mathbf{x}$ 9 0 0

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