This is a question where knowing how to effectively sift through a problem statement and choose only the information you need will really help. We are given a bunch of values, but only need to know one thing, which is that the distance between the two charges is doubled.

Coulomb's law is as follows:

\(\displaystyle F_e= k\frac{q_1q_2}{r^2}\)

We can rewrite this for the initial and final scenarios:

\(\displaystyle F_{ei}= k\frac{q_1q_2}{r_i^2}\)

\(\displaystyle F_{ef}= k\frac{q_1q_2}{r_f^2}\)

We can divide one equation by the other to set up a ratio:

\(\displaystyle \frac{F_{ei}}{F_{ef}} = \frac{r_f^2}{r_i^2}\)

We know that the final radius is double the intial, which is written as:

\(\displaystyle r_f = 2r_i\)

Substituting this in we get:

\(\displaystyle \frac{F_{ei}}{F_{ef}} = \frac{(2r_i)^2}{r_i^2} = 4\)

Rerranging for the final force, we get:

\(\displaystyle F_{ef} = \frac{1}{4}F_{ei}\)

Use Coulomb's law.

\(\displaystyle F=\frac{kq_{1}q_{2}}{r^{2}}\)

Plug in known values and solve.

\(\displaystyle F=\frac{(9\cdot 10^{9})(6\cdot 10^{-6})(9\cdot 10^{-6})}{0.75^{2}}\)

\(\displaystyle F=0.864N\)

Note that a positive value for electric force corresponds to a repulsive force. This should make sense since the charge on both particles are the same sign (positive).

Use Coulomb's law.

\(\displaystyle F=\frac{kq_1q_2}{r^2}\)

\(\displaystyle F=\frac{(9\cdot10^{9})(-6\cdot10^{-6})(-14\cdot10^{-6})}{0.1^{2}}\)

\(\displaystyle F=75.6N\)

Note that the electric force between two charges of the same sign (both positive or both negative) is a positive value. This indicates a repulsive force.

Recall that Coulomb's law tells us that the magnitude of force between two point charges is given as:

\(\displaystyle F=\mid \frac{q_1q_2}{r^2} \mid\)

Here, \(\displaystyle F\) is force between two particles, \(\displaystyle q_1,q_2\) are the charges of each of the two particles, and \(\displaystyle r\) is the distance between the two charges. In our case, \(\displaystyle q_1\) and \(\displaystyle q_2\) are identical since each is the charge of a proton which is given as:\(\displaystyle 1.602*10^{-19}C\), and \(\displaystyle r=3*10^{-9} m\)

Thus, plug in known values and solve.

\(\displaystyle F= \frac{(1.602*10^{-19}C)*(1.602*10^{-19}C)}{(3*10^{-9}m)^2}=\frac{2.556*10^{-38}C^2}{9*10^{-18}m^2}\)

\(\displaystyle F=2.840*10^{-21}N\)

To determine if the force is attractive or repulsive, we only need to examine the sign of the charges. Since both protons have the same sign for their charge (positive charges) they will repel. 

Use Coulomb's law to find the electric force between the charges:

\(\displaystyle F=\frac{kq_1q_2}{r^2}\)

\(\displaystyle F=\frac{9*10^9*1.6*10^{-5}*2*10^{-5}}{0.01^2}\)

\(\displaystyle F=28800N\)

The electric force between two point charges is given by Coulomb's law:

\(\displaystyle F=\frac{kq_{1}q_{2}}{r^2}\)

Now, plug in the given charges (both the same magnitude), the given constant, and the distance between the charges (in meters) to get our answer:

\(\displaystyle F=\frac{9*10^9(3* 10^{-2})^2}{(2* 10^{-9})^2}=2.025*10^{24} N\)

We are given all the necessary information to find the magnitude of the electric force by using Coulomb's law: 

\(\displaystyle \mid\mid F\mid\mid= \mid\frac{kq_1q_2}{r^2}\mid\)

Where \(\displaystyle k\) is Coulomb's constant given by \(\displaystyle 9.0*10^9\frac{N\cdot m^2}{C^2}\), \(\displaystyle q_1\) and \(\displaystyle q_2\) are the respective charges, and \(\displaystyle r\) is the distance between the charges. In our case:

\(\displaystyle \mid\mid F\mid\mid= \mid\frac{9.0*10^9*1*3}{3^3}N\mid=10^9N\)

The method to solving Coulomb's law problems with electrostatic configurations is to find the magnitude of the force and then assign a direction based of what is known about the charges. Coulomb's law is given as:

\(\displaystyle F=k \frac{q_{a} q_{b}}{r^{2}}\)

Where \(\displaystyle q_{a}\) and \(\displaystyle q_{b}\) are the two particles we are finding the force between and \(\displaystyle k\) is the electric constant and is:

\(\displaystyle k=\frac{1}{4 \pi \epsilon _{o}}=8.99\cdot10^{9} \frac{N\cdot m^{2}}{C^{2}}\)

Notice that the distances between \(\displaystyle q_{1}\) and \(\displaystyle q_2\) is the same as the \(\displaystyle q_{2}\) and \(\displaystyle q_{3}\). Since the magnitudes of all charges are the same, that means that the magnitudes of the forces (not directions) are the same. So the force exerted on \(\displaystyle q_{2}\) from \(\displaystyle q_{1}\) is the same magnitude as the force exerted on \(\displaystyle q_{2}\) from \(\displaystyle q_{3}\).

A sketch of the forces is shown below:

Remember that there are always equal and opposite force pairs. We only care about the forces acting on \(\displaystyle q_{2}\) and the last picture shows the two forces that act on it from \(\displaystyle q_{1}\) and \(\displaystyle q_{3}\). Notice that the vector arrows are of equal length (force magnitudes are equal) and in different directions. Coulomb forces obey the law of superposition and we can add them. Before we do that let's calculate the magnitude of the two forces pictured.  

Remember to convert distances to meters and charge magnitudes to Coulombs so the units work out and you are not off by any factors of \(\displaystyle 10\).

\(\displaystyle F=k \frac{q_{a} q_{b}}{r^{2}}=89.9N\)

Both the red vector arrow and the blue vector arrow have magnitudes of \(\displaystyle 89.9N\). Notice in the diagram below that if the charges are spaced equidistant, the will form an equilateral triangle.  

The angle \(\displaystyle \theta=60^{o}\) is the same angle that each vector on the right has relative to the line drawn. In order to add the vectors together we need to separate the components of the vectors into their x- and y-components and add the respective components. This is where symmetry can be handy to make the problem easier. Since the particles are equidistant and the charge magnitudes are all equal, this lead to the force magnitudes to be equal. By inspection it can be shown that the y-components must be equal and opposite and therefore cancel.  

This means that total magnitude of the force acting on \(\displaystyle q_{2}\) is just the sum of the x-component forces. To get the x-components we can use the cosine of the angle. Since the angles are equal and the magnitudes are equal, the final answer will be:

\(\displaystyle F=F_{1}_{x}+F_{2}_{x}=(89.9N)Sin(60^{o})+(89.9N)sin(60^{o})\)

\(\displaystyle \sin(60^{o})=\frac{1}{2}\)

\(\displaystyle F=2(89.9N)\sin(60^{o})=2(89.9N)\left ( \frac{1}{2}\right )=89.9N \hat{x}\)

The final answer is in the positive x-direction, denoted by the positive answer and the \(\displaystyle \hat{x}\) to indicate in the x-direction. The answer must have a magnitude and direction to describe the net force acting on the particle.

From Coulomb's Law: 

\(\displaystyle F=\frac{k*Q*q}{r^2}\)

Where \(\displaystyle r\) is the distance between point charges, \(\displaystyle k=9.0*10^9 \frac{N}{m^2C^2}\), and \(\displaystyle Q\) and \(\displaystyle q\) are charges of the electrons. In our case, \(\displaystyle Q=q=96,000\frac{C}{mol}\). 

\(\displaystyle F=\frac{9*10^9\frac{N}{m^2C^2}*(96,000\frac{C}{mol})^2}{10m^2}= 9.2*10^{17}\frac{N}{mol}\)

First lets set up two axes. Have \(\displaystyle +x\) be to the right of charge 3 and 2 in the diagram and \(\displaystyle +y\) be above charges 1 and 2 in the diagram with charge 2 at the origin.

Coloumb's law tells us the force between point charges is

 \(\displaystyle F_{21}=k\frac{q_{1}q_{2}}{r^{2}}\)

The net force on charge 2 can be determined by combining the force on charge 2 due to charge 1 and the force on charge 2 due to charge 3.

Since charge 1 and charge 2 are of opposite polarities, they have an attractive force; therefore, charge 2 experiences a force towards charge 1 (in the \(\displaystyle -y\) direction). By using Coloumb's law, we can determine this force to be

 \(\displaystyle F_{21}=k\frac{q_{1}q_{2}}{{r_{12}}^{2}}\)

\(\displaystyle F_{21}=(8.99\cdot10^{9}\frac{N\cdot m^2}{C^2})\frac{(3.1\mu C)(1.6\mu C)}{(0.4m)^{2}}=0.279N\) in the \(\displaystyle -y\) direction 

Since charge 2 and 3 have the same polarities, they have a repulsive force; therefore, charge 2 experiences a force away from charge 2 (in the \(\displaystyle -x\) direction). By using Coloumb's law, we can determine this force to be: 

\(\displaystyle F_{23}=k\frac{q_{2}q_{3}}{{r_{23}}^2}\)

\(\displaystyle (8.99\cdot10^{9}\frac{N\cdot m^2}{C^2})\frac{(1.6\mu C)(2.4\mu C)}{(0.3m)^{2}}=0.38357N\) in the \(\displaystyle -x\)-direction

If we draw out these two forces tip to tail, we can construct the net force:

From this, we can see that \(\displaystyle F_{21}\) and \(\displaystyle F_{23}\) create a right triangle with the net force on charge 2 as the hypotenuse. By using the Pythagorean theorem, we can calculate the magnitude of the net force:

 \(\displaystyle F_{2}=\sqrt{{F_{23}}^{2}+{F_{21}}^{2}}=\sqrt{(0.38357N)^{2}+(0.279N)^{2}}=0.474N\)