Answer:
Magnitude: 0.5 N
Direction: 0 degrees
Step-by-step explanation:
Going to be honest, I do not understand the magnitude of R, but what I can explain is the direction.
So, following the steps, we find that the magnitude of A (coulomb's constant k times (q1 x q3)/dq13^2) = 1.798 Newtons
The magnitude of B is the same because q1 = q2 in this scenario
Using SohCahToa, we can find the angle of vector A = about 36.9 degrees
since q1 is 0.3 meters away from the x-axis (this is our opposite side) and the distance between q1 and q3 is 0.5 (our hypotenuse because the x-axis and y-axis make a right angle) this means that we can use this for our formula,
the sine of angle A = opposite side / hypotenuse side
Sine of angle A = 0.3m / 0.5m
Angle A = (inverse Sine) of 0.3m/0.5m
Angle A = 36.9 degrees
Using that we can get the x and y components of each vector:
The x component of A will be A cosine angle A which is
Ax = 1.798 N times (cosine(36.9 degrees)) = 1.44 N
The y component of A will be NEGATIVE A sine angle A
The reason why it is negative is because for the A vector, it has a negative slope so its y value is continually decreasing, so it makes sense for it's resulting vector to have the same y decrease.
Ay = (-1.798)(sine(36.9)) = -1.079 N
Now we can do the same with B.
Since B = A IN THIS SCENARIO, the values will be the same,
Bx = 1.798 N times (cosine(36.9 degrees) = 1.44 N
BUT with the y value here, it is actually INCREASING
so:
By = 1.798 N times (sine(36.9 degrees)) = 1.079 N
So the next step is to sum these values.
Rx = Ax + Bx = 1.44 N + 1.44 N = 2.88 N
The x component of vector R is 2.88 N.
Ry = Ay + By = -1.079 N + 1.079 N = 0 N.
The y component of vector R is 0.
Basically, it is traveling straight along the x-axis. This makes sense because the forces of each particle were repelling this one at an EQUAL MAGNITUDE (q1 = q2) So, the direction is 0 degrees because even following the steps, (tan-1 times (Ry over Rx)) is 0/2.88. This will be 0.