Answer: Option d.
Explanation:
The force between charges can be expressed as:
F = k*q1*q2/r^2
where k is a constant, q1 and q2 are the charges and r is the distance between them.
We hare in a equilateral triangle, so al the distances are equal.
and the charges are:
qa = q
qb = q
qc = 2q
Now, for example, the force that experiments charge A is (in the y axis)
F = (k*qa*qb/r^2 + k*qa*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2
for charge B, the force is again, in the y axis.
F = (k*qb*qa/r^2 + k*qb*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2
for particle C, we have:
F = (k*qc*qa/r^2 + k*qc*qb/r^2)*cos(30°) = cos(30°)*(q + q)*k*2q/r^2 = cos(30°)*4*k*q^2/r^2
Wher the cosine of 30° comes because we have a equilatiral triangle, where all the internal angles are 60°, so if we draw a line that cuts the angle by half (our y-axis) the angles to each side are 30°.
We can do a similar process for the forces in the x-axis, and we will reach the same conclusion:
Now, this means that the force that experiences the charge C is the biggest force, so the correct option is c.
