Question

An equilateral triangle with side lengths of 0.50 m has a 5.0 nC charge placed at each corner. What is the magnitude of the electric field at the midpoint of one of the three sides? (A) 240 N/C (B) 180 N/C (C) 720 N/C (D) 480 N/C (E) 120 N/C

Answer 1

To solve this problem we will first find the distance between each of the points 'x' and then use it as the variable of the distance in the function of the electric field. According to the graph the value of 'x' is,

`x = (√(3))/(2)a`

`x = (√(3))/(2)(0.5)`

`x = 0.43301m`

The magnitude of the electric field is

E=\frac{kQ}{x^2}

Here,

k = Coulomb's Constant

Q = Charge

x = Distance

`E=((9*10^9)(5*10^(-9)))/((0.43301)^2)`

`E = 240.002N/C \approx 240N/C`

Therefore the correct answer is A.