In this problem we have charge $q_1 = – 2 cdot 10^{-4 } ~mathrm{C}$ and charge
$q_2 = 8 cdot 10^{-4 } ~mathrm{C}$ at a distance $r = 0.3 ~mathrm{m}$ from each other. To find the force between these two charges, we’ll use the following equation for force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other:

$$
F = dfrac{k q_1 q_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant. When we plug in the given values we have:

$$
begin{align*}
F &= dfrac{k q_1 q_2}{r^2} \
F &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot ( – 2) cdot 10^{-4 } ~mathrm{C} cdot 8 cdot 10^{-4 } ~mathrm{C} }{(0.3 ~mathrm{m})^2 } \
F&= dfrac{- 144 cdot 10^2 ~mathrm{N}}{0.09}
end{align*}
$$

$$
boxed{F = – 160 000 ~mathrm{N}}
$$

$$
F = – 160 000 ~mathrm{N}
$$

In this problem we have a negative charge $q_1 = -6 cdot 10^{-6} ~mathrm{C}$ exerting an attractive force of $|F_e| = 65 ~mathrm{N}$ on an unknown charge $q_2$ that’s at a distance $r = 0.05 ~mathrm{m}$ from charge $q_1$.
Electric force $F_e$ between charges $q_1$ and $q_2$ at a distance $r$ from each other is given as:

$$
begin{equation}
F = dfrac{k q_1 q_2}{r^2}
end{equation}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.

We’re told that this force is attractive, which means that magnitude of the force $F_e$ is negative and charge $q_2$ is positive.
Let’s apply equation $(1)$ to a system of these two charges:

$$
begin{align*}
F_e &= dfrac{k q_1 q_2}{r^2} \
tag{multiply by $r^2$} \
F_e r^2 &= k q_1 q_2 \
tag{express $q_2$ from the equation above} \
q_2 &= dfrac{F_e r^2 }{k q_1} \
tag{plug in the given values. Note that $F_e = – 65 ~mathrm{N} $} \
q_2 &= dfrac{ – 65 ~mathrm{N} cdot (0.05 ~mathrm{m})^2 }{9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot (-6) cdot 10^{-6} ~mathrm{C} } \
q_2 &= dfrac{ 65 ~mathrm{N} cdot 2.5 cdot 10^{-3} ~mathrm{m}^2 }{9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 6 cdot 10^{-6} ~mathrm{C} }
end{align*}
$$

$$
boxed{ q_2 = 3 cdot 10^{-6 } ~mathrm{C} }
$$

$$
q_2 = 3 cdot 10^{-6 } ~mathrm{C}
$$

Three charges $q_{A}= 6 mathrm{~mu C}$, $q_{B}= 3 mathrm{~mu C}$ and $q_{C}= 1.5 mathrm{~mu C}$ are placed like shown in the figure. Distance between charges $q_{A}$ and $q_{B}$ is $r_{AB}=4 mathrm{~cm}$ whereas distance between charges $q_{A}$ and $q_{C}$ is $r_{AC}= 3 mathrm{~cm}$. All three charges have the same sign, and thus repel each other. We need to determine net force that acts on charge $q_{A}$.
Net force that acts on charge $q_{A}$ is equal to:

$$
vec{F}_{net}=vec{F}_{BA}+vec{F}_{CA}
$$

where $vec{F}_{BA}$ is force between charges $q_{B}$ and $q_{A}$, $vec{F}_{CA}$ is force between charges $q_{C}$ and $q_{A}$. Magnitude of net force on charge $q_{A}$ can be determined by applying Pythagorean theorem on triangle formed by vectors $vec{F}_{BA}$, $vec{F}_{CA}$ and $F_{net}$. We find that magnitude of $F_{net}$ is equal to:

$$
F_{net}=sqrt{(F_{BA})^{2}+(F_{CA})^{2}}
$$

Magnitude of electric force between charges $q_{B}$ and $q_{A}$ is equal to:

$$
F_{BA}=kcdot dfrac{q_{A}cdot q_{B}}{r_{AB}^2}
$$

where $k=9cdot 10^{9} mathrm{~dfrac{Ncdot m^2}{C^2}}$ is Coulomb’s constant, and $r_{AB}^{2}$ is distance between charges $q_A$ and $q_B$.

$$
begin{align*}
F_{AB}&= 9 cdot 10^{9} mathrm{~dfrac{Ncdot m^2}{C^2}}
cdot dfrac{6 mathrm{~mu C} cdot 3 mathrm{~mu C}}{(4 mathrm{~cm})^{2}}\
tag{ $1 ~mathrm{mu C } = 10^{-6} ~mathrm{C}$} \
tag{$1 ~mathrm{cm} = 0.01 ~mathrm{m}$ } \
F_{AB} &= 9 cdot 10^{9} mathrm{~dfrac{Ncdot m^2}{C^2}}
cdot dfrac{6 cdot 10^{-6} mathrm{~C} cdot 3 cdot 10^{-6} mathrm{~C}}{(0.04 mathrm{~m})^{2}}
end{align*}
$$

$$
boxed{F_{AB} = 101.25 mathrm{~N}}
$$

Analogously, magnitude of electric force between charges $q_{C}$ and $q_{A}$ is equal to:

$$
begin{align*}
F_{CA}&=kcdot dfrac{q_{A}cdot q_{C}}{r_{AC}^2}\
F_{CA}&= 9 cdot 10^{9} mathrm{~dfrac{Ncdot m^2}{C^2}} cdot dfrac{6 mathrm{~mu C} cdot 1.5 mathrm{~mu C}}{(3 mathrm{~cm})^{2}}\
tag{ $1 ~mathrm{mu C } = 10^{-6} ~mathrm{C}$} \
tag{$1 ~mathrm{cm} = 0.01 ~mathrm{m}$ } \
F_{CA}&= 9 cdot 10^{9} mathrm{~dfrac{Ncdot m^2}{C^2}} cdot dfrac{9 cdot 10^{-6} mathrm{~C}}{(0.03 mathrm{~m})^{2}}
end{align*}
$$

$$
boxed{F_{CA} =90 mathrm{~N}}
$$

We can plug in the calculated values for magnitude of force $F_{AB}$ and $F_{AC}$ into equation for magnitude of net force acting on charge $q_{A}$:

$$
begin{align*}
F_{net}&=sqrt{(F_{BA})^{2}+(F_{CA})^{2}}\
F_{net}&=sqrt{(101.25 mathrm{~N})^{2}+(90 mathrm{~N})^{2}}\
F_{net}&=sqrt{10251.56 mathrm{~N}+8100 mathrm{~N}}
end{align*}
$$

$$
boxed{F_{net} =135.46 mathrm{~N}}
$$

We calculated magnitude the net force acting on charge $q_{A}$, but we also need to find its direction. From the figure:

$$
begin{align*}
tan(Theta)&=dfrac{F_{CA}}{F_{BA}}\
Theta&=arctan left(dfrac{F_{CA}}{F_{BA}}right)\
Theta&= arctan left(dfrac{90 mathrm{~N}}{101.25 mathrm{~N}}right)
end{align*}
$$

$$
boxed{Theta = 41.63^{circ}}
$$

Hint: Find the magnitude of the net force $F_{net}$ acting on charge $A$ and angle at which this force acts on charge $A$.

$$
F_{net} =135.46 mathrm{~N}
$$

$$
Theta = 41.63^{circ}
$$

In this problem we have a positive charge $q_A = 2 cdot 10^{-6} ~mathrm{C}$ placed at the origin.
Negative charge $q_B = – 3.6 cdot 10^{-6} ~mathrm{C}$ is at a coordinate $x_B = r_B = 0.6 ~mathrm{m}$. This coordinate also represents the distance $r_B$ between this charge and charge $A$.
Positive charge $q_C = 4 cdot 10^{-6} ~mathrm{C}$ is at a coordinate $x_C= r_C= 0.8 ~mathrm{m}$ from charge $A$. This coordinate also represents the distance $r_C$ between this charge and charge $A$.

Net force $F$ acting on charge $A$ is equal to a sum of individual forces acting on charge $A$, one being the electrostatic force $F_B$ between charge $A$ and charge $B$ and the other one being electrostatic force $F_C$ between charge $A$ and charge $C$:

$$
begin{equation}
F = F_B + F_C
end{equation}
$$

Notice that charge $A$ is repelled from charge $C$, so magnitude of the force between them is positive. Charge $A$ is attracted to charge $B$, so magnitude of the force between charges $A$ and $B$ is negative.
We’ll use the following equation to calculate the force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other:

$$
F = dfrac{k q_1 q_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.

When we plug in the corresponding values for distance $r_B$ between charges $A$ and $B$ and values for charges $q_A$ and $q_B$ we find that force $F_B$ between these two charges is equal to:

$$
F_B = dfrac{kq_A q_B}{r_B^2}
$$

When we plug in the corresponding values for distance $r_C$ between charges $A$ and $C$ and values for charges $q_A$ and $q_C$ we find that force $F_C$ between these two charges is equal to:

$$
F_C = dfrac{kq_A q_C}{r_C^2}
$$

We can plug in the conclusions stated above into equation $(1)$ to find net force acting on charge $A$:

$$
begin{align*}
F &= F_B + F_C \
F &= dfrac{kq_A q_B}{r_B^2} + dfrac{kq_A q_C}{r_C^2} \
F &= kq_A bigg( dfrac{ q_B}{r_B^2} + dfrac{ q_C}{r_C^2} bigg) \
F &= 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 2 cdot 10^{-6} ~mathrm{C} cdot bigg( dfrac{ – 3.6 cdot 10^{-6} ~mathrm{C} }{(0.6 ~mathrm{m})^2} + dfrac{ 4 cdot 10^{-6} ~mathrm{C}}{(0.8 ~mathrm{m})^2} bigg) \
F&= 18 cdot 10^3 cdot 10^{-6} cdot
bigg( dfrac{ – 3.6 }{0.36 } + dfrac{ 4 }{0.64 } bigg) ~mathrm{ N}
end{align*}
$$

$$
boxed{ F = – 0.0675 ~mathrm{ N} }
$$

$$
F = – 0.0675 ~mathrm{ N}
$$

In this problem we have a positive charge $q_A = 2 cdot 10^{-6} ~mathrm{C}$ placed at the origin.
Negative charge $q_B = – 3.6 cdot 10^{-6} ~mathrm{C}$ is at a coordinate $x_B = r_A = 0.6 ~mathrm{m}$. This coordinate also represents distance between charge $q_A$ and $q_B$.
Positive charge $q_C = 4 cdot 10^{-6} ~mathrm{C}$ is at a coordinate $x_C = 0.8 ~mathrm{m}$.

Net force $F$ acting on charge $B$ is equal to a sum of individual forces acting on charge $B$, one being the electrostatic force $F_A$ between charge $A$ and charge $B$ and the other one being electrostatic force $F_C$ between charge $B$ and charge $C$:

$$
begin{equation}
|F| = |F_A| – |F_C|
end{equation}
$$

Notice that charge $B$ is attracted by charge $C$, so magnitude of the force between them is negative. Charge $B$ is also attracted to charge $A$, so magnitude of the force between charges $A$ and $B$ is negative.
We’ll use the following equation to calculate the force $F$ between charges $q_1$ and $q_2$ at a distance $r$ from each other:

$$
F = dfrac{k q_1 q_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.

When we plug in the corresponding values for distance $r_A$ between charges $A$ and $B$ and values for charges $q_A$ and $q_B$ we find that force $F_A$ between these two charges is equal to:

$$
F_A = dfrac{kq_A q_B}{r_A^2}
$$

In order to use the general equation above to find the force between charge $q_B$ and charge $q_C$, we first need to find the distance between these two charges. Distance $r_C$ between charge $q_B$ and charge $q_C$ is equal to:

$$
r_C = x_C – x_B = 0.8 ~mathrm{m} – 0.6 ~mathrm{m} = 0.2 ~mathrm{m}
$$

When we plug in the corresponding values for distance $r_C$ between charges $B$ and $C$ and values for charges $q_B$ and $q_C$ we find that force $F_C$ between these two charges is equal to:

$$
F_C = dfrac{kq_B q_C}{r_C^2}
$$

We can plug in the conclusions stated above into equation $(1)$ to find net force acting on charge $q_B$:

$$
begin{align*}
F &= F_A + F_C \
F &= dfrac{kq_A q_B}{r_A^2} – dfrac{kq_B q_C}{r_C^2} \
F &= kq_B bigg( dfrac{ q_A}{r_A^2} – dfrac{ q_C}{r_C^2} bigg) \
F &= 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot (- 3.6) cdot 10^{-6} ~mathrm{C} cdot
bigg( dfrac{ 2 cdot 10^{-6} ~mathrm{C} }{(0.6 ~mathrm{m} )^2} – dfrac{ 4 cdot 10^{-6} ~mathrm{C} }{( 0.2 ~mathrm{m} )^2} bigg) \
F&= – 32.4 cdot 10^3 cdot 10^{-6} cdot
bigg( dfrac{ 2 }{0.36 } – dfrac{ 4 }{0.04 } bigg) ~mathrm{ N}
end{align*}
$$

$$
boxed{ F = 3.06 ~mathrm{ N} }
$$

Note that since we assumed the magnitude of force $F$ to be calculated as
$|F| = |F_A| – |F_C|$ and we calculated the positive value of force $F$, this means that magnitude of force $F_A$ is greater than magnitude of force $F_C$, which means that direction of force $F$ is from charge $B$ to charge $A$.

$$
F = 3.06 ~mathrm{ N}
$$