Three point charges of 2microcoulomb,-3microcoulomb,and -3microcoulomb are kept at the vertices A, B and C respectively of the equilateral triangle of side 20centimeter . What should be the sides of the triangles and magnitude of the charge to be place at the mid point M of the side BC so that the charge at A remains equilibrium

Answer 1

the line from A to the midpoint is forms a perpendicular with BC, so the length of that line is sqrt(20^2-10^2)=sqrt300

Now, considering the forces of attration from B or C, the portion parallel to the Midpoint line, will be (k(-2*3)/20^2 )*cos30 for each charge at B and C, but since those forces are both attractive in a line parallel to the Midpoint line, they add.
Net force= -3*2*2k*cos30/20^2
but you want this force cancled by new charge Q
force=k*Q/distance=kQ/(20^2-10^2)
adding the new force and old net force and setting to zero, we can get
kQ/300=12kcos30/400
so solve for Q
In my head I get Q=9*.866 microC

Answer 2

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