Question

Three charges are fixed in the x?y plane as follows: 1.3nC at the origin (0, 0); 2.6nC at (0.60m , 0);

Answer 1

Cahrges, positions and direction of the force:

q₁ = 1.3 * 10^-9 C           at Origin (0, 0)                   Where force need to be calculated

q₂ = 2.6 * 10^-9 C           at Origin (0.6 m, 0)            Force along -x - axis and contributes only to the x component of net force.

q₃ = -2.1 * 10^-9 C          at Origin (0, 1.7 m)            Force along +y - axis and contributes only to the y component of net force.

X component of the force acting on q₁ due to q₂ and q₃ respectively is,

F_x = k q₁ q₂ cos 180 / (0.6)² + k q₁ q₃ cos 90 / (1.7)²                 where k = 8.99 * 10⁹ N.m² / C² is the Coulomb's force constant.

F_x = 8.99 * 10⁹ * 1.3 * 10^-9 * 2.6 * 10^-9 * (-1)/ (0.6)² + 0

F_x = - 84.406 * 10^-9 N

Y component of the force acting on q₁ due to q₂ and q₃ respectively is,

F_y = k q₁ q₂ sin 180 / (0.6)² + k q₁ q₃ sin 90 / (1.7)²

F_y = 0 + 8.99 * 10⁹ * 1.3 * 10^-9 * 2.1 * 10^-9 * (1)/ (1.7)²

F_y = 8.492 * 10^-9 N

Net force magnitude is,

F = (F_x² + F_y²)^1/2

F = 84.832 * 10^-9 N

Answer 2

let q1 = 1.3 nc, q2 = 2.6 nc and q3 = -2.1 nc

F12 = k*q1*q2/x^2

= 9*10^9*1.3*2.6*10^-18/0.6^2

= 0.845*10^-9 N

F13 = k*q1*q3/y^2

= 9*10^9*1.3*2.1*10^-18/1.7^2

= 8.5*10^-9 N

Fnet on q1 = sqrt(F12^2 + F13^2)

= sqrt(0.845^2+8.5^2)*10^-9

= 8.54*10^-9 N