should be located in between them (so that the field vectors point in the opposite direction). Let the coordinate of the second particle be x'(x'> 0). Then, the magnitude of the field due to the charge –q1evaluated at xis given by E= q1/4πε0x2, while that due to the second charge –4q1is E'= 4q1/4πε0(x'– x)2. We set the net field equal to zero: GEEEnet=¡=′0 so that qxqxx102102444ππε=′ −bg. Thus, we obtain x'= 3x= 3(2.0 mm) = 6.0 mm.
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Electric charge, Electric field vectors, negative x direction, field vectors