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Originally Posted by threewingedfury
so you do (1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2) for particle 1 and give it an 'i' and -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2) for particle 2 and give it a j?
So I dont need to use the -5 microcoulombs? keep it positive?
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you need to calculate x- and y-components for EACH charge, so in total you'll have 4 uses of Coulomb's Law. The signs are tricky, you do need to keep track of positive and negative charge, but I prefer doing the number crunching in absolute value and applying the signs later. In this case, the field resulting from particle 1 has a positive x-component and a negative y-component because field lines point from positive charge to negative charge. If you draw a diagram with field lines that should be more clear.