Question

Three identical point charges of charge q = 5 uC are placed at the vertices (corners) of an equilateral triangle. If the side of triangle is a = 3.3m, what is the magnitude, in N/C, of the electric field at the point P in one of the sides of the triangle midway between two of the charges?

Answer 1

Electric field due to a charge Q at a distance r is

E = k Q / r²            where k = 8.99 * 10⁹ N.m² /C² is the Coulomb's force constant.

Let's say, P is the mid point of base of the triangle with a = 3.3 m and thus,

q₁ = 5 * 10^-6 C on the lower left vertex and r = 1.65 m

q₂ = 5 * 10^-6 C on the lower right vertex and r = 1.65 m

q₃ = 5 * 10^-6 C on the lower left vertex and r = 2.858 m

Now,

X component of the Electric field at point P due to q₁, q₂, and q₃ respectively is,

E_x = k q₁ cos0 / (1.65)² + k q₂ cos180 / (1.65)² + k q₃ cos270 / (2.858)²

E_x = 8.99 * 10⁹ * 5 * 10^-6 [(1-1)/(1.65)² + 0]

E_x = 0 N/C

Y component of the Electric field at point P due to q₁, q₂, and q₃ respectively is,

E_y = k q₁ sin0 / (1.65)² + k q₂ sin180 / (1.65)² + k q₃ sin270 / (2.858)²

E_y = 8.99 * 10⁹ * 5 * 10^-6 [0+ 0 + (-1)/(2.858)²]

E_y = - 5503.07 N/C

Thus, magnitude of the electric field at point P midway between two base vertex charges is,

E = ( E_x)² + ( E_y)²

E = 5503.07 N/C