I\'ve managed to figure out the first two parts, but I\'m stuck on the third. Th

Question

I've managed to figure out the first two parts, but I'm stuck on the third. The trivial solution to C is the center of the triangle, but there are three more locations along the lines of symmetry the triangle has. Essentially these are all 120 degrees apart so you only really need to find one. I picked the on where x = 0 and it is above y = 0. I am struggling to find out the exact point, but i do know that the forces in the y direction cancel.

Problem 12. Consider an equilateral triangle with sides of length s & with equal point charges +Q situated at each of the vertices, as illustrated. Choose the origin of coordinates O = (0.0) at the center of the triangle with x, y-coordinate axes in the plane of the triangle. (a) Determine the electrostatic potential V(x.y) in the x, y-plane. (b) Determine the electrostatic field E(x,y) =-Vfx.y) (c) Determine the coordinate positions of all of the points in the finite x, y-plane at which the electrostatic field vanishes. Provide a physical explanation why E = 0 at each such point

Explanation / Answer

given an equilateral triangle with charge +Q on its vertices

side of equilateral triangle = s

and origin is at the center of the triangle

so the coordinates of the three charges are

(0.5s, -0.2886s)

(-0.5s, -0.2886s)

(0, 0.5773s)

a. so at any point (x,y)

potential is given by

V = kQ/sqroot[(x - 0.5s)^2 + (y + 0.2886s)^2] + kQ/sqroot[(x + 0.5s)^2 + (y + 0.2886s)^2] + kQ/sqroot[(x)^2 + (y - 0.5773s)^2]

V = kQ[1/sqroot[(x - 0.5s)^2 + (y + 0.2886s)^2] + 1/sqroot[(x + 0.5s)^2 + (y + 0.2886s)^2] + 1/sqroot[(x)^2 + (y - 0.5773s)^2]]

b. similiarly

electric field at any point (x,y) is given by

E = kQ[(x - 0.5s)i + (y + 0.2886s)j]/[(x - 0.5s)^2 + (y + 0.2886s)^2]^3/2 + kQ[(x + 0.5s)i + (y + 0.2886s)j]/[(x + 0.5s)^2 + (y + 0.2886s)^2]^3/2 + kQ[(x)i + (y -0.5773s)j]/[(x)^2 + (y - 0.5773s)^2]^3/2

E = kQ{[(x - 0.5s)i + (y + 0.2886s)j]/[(x - 0.5s)^2 + (y + 0.2886s)^2]^3/2 + [(x + 0.5s)i + (y + 0.2886s)j]/[(x + 0.5s)^2 + (y + 0.2886s)^2]^3/2 + [(x)i + (y -0.5773s)j]/[(x)^2 + (y - 0.5773s)^2]^3/2}

c. now, for E = 0, Ex = 0 and Ey = 0

as the algebraic expression would be difficult to solve

the intersection of the three circles of equal radius, r drawn with centers on the three charges should have electric field 0 as the vectors due to individual filds of each particle would cancel each other out at these points. Equal radius r because the charges are the same at the three vertices

now, from properties of circle, three circles at three vertices of an equilateral triangle will intersect only at one point for a radius r = 0.5773s ( where s = side of the equilateral triangle) and hence there is just one point where electric field will be 0, and that is the origin

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