What is the formula for determining the resultant force on a charge?

[t_n_p]

Homework Statement

1. A charge A of -5 µC has two nearby charges: +3.0 µC at a distance 0.3 m vertically above it, and +4.0 µC at a distance of 0.4 m horizontally. Find the magnitude of the force in Newton on charge A. Hint: find the vertical and horizontal components, then add these vectors to obtain the resultant.

2. Two charges of +1 and +36 nanocoulomb are separated by a distance of 100 mm. A 3rd test charge is moved along the line between these two charges. Where is the test charge when it experiences a net force = zero?

Give your answer in mm measured from the first of the two charges. Hint: think about the ratio of the two charges and deduce a ratio for the distances.
The attempt at a solution

1. Using the formula |F| = k[(|q1||q2|)/r^2)]

I found Fhorizontal = 1.12*10^-6
and Fverticle = 1.5*10^-6

I then use pythag, resultant = sqrt( (F_horizontal)^2 + (F_vertical)^2 )
to give answer of 1.87*10^-6.
I think it may be right, can somebody verify?

2. I'm not sure how to do this one!

 

[chaoseverlasting]

1. Havent checked the values, but the method looks right.

2. Let the distance from the charges at the point where Fnet is zero be (100-x) and x in mm. Now, find the force exerted by each of the charges on your moving charge and equate them. You should get a quadratic in x. Solve for x.

 

[t_n_p]

Managed to do those ones, but I've got a new question!

Calculate the resultant force in Newton on the left hand charge q = 12 µC due to charges q = -3 µC at distance 0.1 m to the right and a charge 13 µC at a further distance 0.2 m to the right (measured from the 2nd charge).
Use the correct sign, taking force to the right as positive and to the left as negative.
12 µC......q =-3 µC....... 13 µC

|<------ 0.1 m ----->|<------------ 0.2 m-------------->|

A step by step guide would be very handy! Thanks in advance to anybody who can help

 

[chaoseverlasting]

Give it a shot first. Use coulombs law and make a FBD to point yourself in the right direction.