What is the electric field strength at the center of the triangle?

Question

Three 19-cm-long rods form an equilateral triangle. Two of the rods are charged to + 15 nC, the third to - 15 nC.

What is the electric field strength at the center of the triangle?

Answer 1

♠take origin of coordinate system (0,0,0) at the center of our triangle;  

half of a rod’s length is b=0.5*19 cm = 0.095 m;

according to properties of equilateral triangle the distance of a midpoint of a rod is a=b/√3;

♦lets consider first a rod parallel to x-axis; let the midpoint of a rod with charge q have coordinates (0, a), ends being (-b, a) and (b, a);

♦elementary strength of electric field produced by elementary charge dq in the triangle center (0,0,0) is a vector

dE = -dq*r /(4pi*ε0*|r|^3), where ε0=8.854e-12,

vector r=(x, a) is position of elementary charge dq=p*dx,

p=q/(2b) is linear charge density,

|r|=√(x^2 +a^2) is distance of elementary charge dq from (0,0,0)  

♦ strength of electric field produced by the rod above is vector

E= -m*∫dx*(x, a)/√(x^2 +a^2)^3 {x=-b to b}, where m=p/(4pi*ε0);

∫dx*x/√(x^2 +a^2)^3 = -1/√(x^2 +a^2) = 0; as expected!

∫dx*a/√(x^2 +a^2)^3 = (x/a) /√(x^2 +a^2) =

= (2b/a)/√(b^2 +a^2) = (2b/a)/(2*a) = 3/b;

vector E = (0, -3m/b);  

♣ 3 rods: total strength of electric field produced is

vector T=E+E1+E2, where E is found above,

vector E1= (3*m1/b) *(-0.5*√3, 0.5),  

vector E2= (3*m2/b) *(+0.5*√3, 0.5), m1=m2 =-m;  

vectors E1 and E2 are got by rotating E by angles  Ã‚±120°;

T=(3m/b) *(0, 2) =

= (0, 3*15e-9/(4pi *8.854e-12 *0.095^2)) =(0, 44814.23) N/C;