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.

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
In this answer where do the final values of 5.1961x10-9nC come from i follow until the last step starting with q/v3=3e-9

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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