Combine Newton's second law with the equation for electric force due to an electric field:

$\displaystyle F_{net}=ma=F_E=qE$

Plug in values:

$\displaystyle 25*2.1=4.5q$

$\displaystyle q=11.66C$

Since the electric field is pointing towards the charge, it is known that the charge has a negative value.

Using electric field formula:

$\displaystyle |E|=k\frac{|q|}{r^2}$

Solving for $\displaystyle q$

$\displaystyle |q|=|E|\frac{r^2}{k}$

Plugging in values:

$\displaystyle q=15*\frac{2^2}{9*10^9}$

$\displaystyle |q|=6.67nC$

Since the charge must have a negative value:

$\displaystyle q=-6.67nC$

The equation for the force experienced by two point charges is known as Coulomb's Law, and is as follows.

$\displaystyle F=\frac{kq_1q_2}{r^2}$

The value 'k' is known as Coulomb's constant, and has a value of approximately $\displaystyle 9.0 \times 10^{9} N$.

We have all of the numbers necessary to use this equation, so we can just plug them in.

$\displaystyle \begin{align*} F&= \frac{kq_1q_2}{r^2}\\ &= \frac{(9\times 10^9)(5\times 10^{-6})(-2\times 10^{-6})}{(5^2)} \\ &= -0.0036\\ &= -3.6\times 10^{-3} \end{align*}$

Since we're given a negative number (and through our intuition: "opposites attract"), we can determine that the force is attractive. Because we're asked for the magnitude of the force, we take the absolute value, so our answer is 

$\displaystyle 3.6 \times 10^{-3} \textup{ N}$, attractive force.

To find the strength of an electric field generated from a point charge, you apply the following equation.

$\displaystyle E=\frac{k*Q}{r^2}$

We know the value of Q and r (the charge and distance, respectively), so we can simply plug in the numbers we have to find the answer.

$\displaystyle \begin{align*} E&= \frac{9E9*10E{-6}}{3^2}\\ &= 10000\ \frac{N}{C} \end{align*}$

While this might seem like a very large number coming from such a small charge, remember that the typical charges interacting with it will be in the same magnitude of strength, roughly. This yields a force much smaller than 10,000 Newtons.

First, let's calculate the electric field at C due to point A.

$\displaystyle E_{1}=k\frac{Q_{1}}{R^2}$

$\displaystyle E_{2}=k\frac{Q_{2}}{R^2}$

We can tell that the net electric field will be in the $\displaystyle -x$ direction.

$\displaystyle E_{net}=E_{2}-E_{1}$

$\displaystyle E_{net}=\frac{k\cdot 10^{-9}}{1^2}(3-\frac{1}{2})$

$\displaystyle E_{net}= 22.5 \frac{N}{C}$ in the $\displaystyle -x$ direction.

Use the equation to find the magnitude of an electric field at a point.

$\displaystyle E_{field}=\frac{kq}{r^2}$ 

$\displaystyle E_{field}=9\cdot10^{9}\frac{N\cdot m^{2}}{C^{2}}\cdot\frac{15mC}{(15m)^{2}}$

Solve.

$\displaystyle E_{field}=600000=6\cdot 10^5\frac{N}{C}$

Since it is a positive charge, the electric field lines will be pointing away from the charged particle.

We will need to use the electric field equation, twice. Because we are given coordinates, we will need to use vector notation.

$\displaystyle E_{total}=E_{1}+E_{2}$

$\displaystyle E_{1}=k\frac{q_{1}}{r_{1}^2} \widehat{r_1}$

$\displaystyle E_{2}=k\frac{q_{2}}{r_{2}^2} \widehat{r_{2}}$

Combine the two equations.

$\displaystyle E_{total}=k\frac{q_{1}}{r_{1}^2} \widehat{r_1} + k\frac{q_{2}}{r_{2}^2} \widehat{r_2}$

Plug in known values. 

$\displaystyle 0 = \frac{50 nC}{5^2}\frac{< 0, 5>}{5}+ \frac{q_2}{8^2}\frac{< 0, 8>}{8}$ 

$\displaystyle q_2 = 128 nC$

Note that the charge is positive. This is because the electric field lines point towards the negative charge at the origin, and in order to balance this at your location, the electric field lines of the charge at (0,-3) must be pointing away from the charge.

Potential is not a vector, so we just add up the two potentials and set them to each other. The equation for electric potential is:

$\displaystyle V=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r}$

$\displaystyle \frac{1}{4\pi \epsilon _{0}}\frac{Q_A}{d}=\frac{1}{4\pi \epsilon _{0}}\frac{Q_B}{(2-d)}$ 

If the point we are looking for is distance $\displaystyle d$ from $\displaystyle Q_A$, it's $\displaystyle (2-d)$ from $\displaystyle Q_B$. Cancel all the common terms, then cross-multiply:

$\displaystyle \frac{1}{4\pi \epsilon _{0}}\frac{3\times10^{-9}}{d}=\frac{1}{4\pi \epsilon _{0}}\frac{2\times10^{-9}}{(2-d)}$

$\displaystyle \frac{3}{d}=\frac{2}{(2-d)}$

$\displaystyle 3(2-d)=2d$

$\displaystyle d=1.2 m$ 

Since we had $\displaystyle d$ associated with $\displaystyle Q_A$, it's from that charge toward the weaker charge.

Electric field is a vector. In between the charges is where $\displaystyle Q_A$'s field points right and $\displaystyle Q_B$'s field points left, so somewhere in between, the two vectors will add to zero. It will be closer to the weaker charge, $\displaystyle Q_B$, but since field depends on the inverse-square of the distance, it will not be linear, and we'll have to do some math.

First set the magnitudes of the two fields equal to each other. The vectors point in opposite directions, so when their magnitudes are equal, the vector sum is zero.

$\displaystyle E_A=E_B \newline \frac{1}{4\pi _0}\frac{Q_A}{d_A^2}=\frac{1}{4\pi _0}\frac{Q_B}{d_B^2}\newline \frac{1}{4\pi _0}\frac{5\times10^{-9}}{d_A^2}=\frac{1}{4\pi _0}\frac{3\times10^{-9}}{(1-d_A)^{2}} \newline \frac{5}{d_A^{2}}=\frac{3}{(1-d_A)^{2}}$

Many of the terms cancel, making it a bit easier. Now cross multiply and solve the quadratic:

$\displaystyle 2d_A^{2}-10d_A +5=0$

$\displaystyle d_A=0.56 m$

Using coulombs law to solve

$\displaystyle F=k\frac{q_1q_2}{r^2}$

 Where:

$\displaystyle q_1$ it the first charge, in coulombs.

$\displaystyle q_2$ is the second charge, in coulombs.

$\displaystyle r$ is the distance between them, in meters

$\displaystyle k$ is the constant of $\displaystyle 9*10^9\frac{N*m^2}{C^2}$

Converting $\displaystyle cm$ into $\displaystyle m$

$\displaystyle 50cm*\frac{1m}{100cm}=.50m$

Plugging values into coulombs law

$\displaystyle F=9*10^{9}\frac{-9*10^{-13}8*10^{-13}}{.50^2}$

$\displaystyle F=-2.59*10^{-14}N$

Magnitude will be the absolute value

$\displaystyle |F|=2.59*10^{-14}N$