posted by Anonymous .
three point charges-1*10^-6,-2*10^-6,3*10^-6.one kept at corners of an equilateral triangle of side 1m.what is potential at centre
The distance from a corner of the triangle to the center is 1/4 the side length, or 0.25 m.
Add up -k*(-1 -2 +3)*10^-6/(0.25)
where k is the Coulomb constant. The sum is zero.
let the center of the triangle be (0,0) and let one vertex be at (1/√3,0). Then the other vertices are at vectors (1/√3,2π/3) and (1/√3,4π/3)
A good discussion of this kind of problem can be found at
I got the distance from the center to the corner wrong; it is (2/sqrt3)*(1/2) = 1/sqrt3 meters. Because the distance is the same for each corner, you add up the charges in the numerator to get the total potential. The answer is still zero.