Point charges of +2.00 nC and - 5.00 nC are arranged on two of the vertices of a

Question

Point charges of +2.00 nC and - 5.00 nC are arranged on two of the vertices of an equilateral triangle. Find the magnitude and sign of the third charge which must be placed at the 3rd vertex of the triangle so that the potential at a point on the side midway between the +2.00 nC and - 5.00 nC is equal to zero. You do not need to know the size of the triangle.

Can somebody please show me how to work this problem from start to finish? I understand the basics behind this but need a little help to make sure I am using the correct procedures. Thanks.

Explanation / Answer

S(k*q/r) = 0 as electric potential is scalar quantity. so, k* S(q/r) = 0 as k is constant so, S(q/r) = 0 let third charge be q let side of triangle be a so, distance of 2.00 nC and - 5.00 nC from thier midpoint = a/2 also, distance of third vertex from that point = v3*a/2 inputing values:- q/(v3*a/2) + (2 e-9)/(a/2) + (-5 e-9)/(a/2) = 0 multiplying throughout by (a/2) (to get rid of it in denominator) and solving for q:- q/v3 = 3 e-9 so, q = v3 * 3 e-9 q = 5.1961 x 10-9 C ˜ 5.20 nC

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