Answer:
electric field Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]
Explanation:
The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.
a) Let's write the electric field for each charge and the total field
E = k q /r
With k the Coulomb constant, q the charge and r the distance of the charge to the test point
Et = E1 + E2 + E3
E1 = k q / (x-a)²
E2 = k (-2q) / x²
E3 = k q / (x + a)²
Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]
The direction of the field is along the x axis
b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ where x << 1, for this we take factor like x from all the equations
Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]
We use binomial expansion
(1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1
(1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...
They replace in the total field and leaving only the first terms
Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]
Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]
Et = k q 2a²/x⁴
point charge
Et = k q 1/x²
Dipole
E = k q a/x³
