1. Given data

• Charges on the $x$-axis:
  • At $x=0$: $q$
  • At $x=\frac{3R}{4}$: $-2q$
  • At $x=R$: $2q$
• $q = 2 \times 10^{-6}\,\text{C}$
• $R = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m}$
• Coulomb constant: $k = 9 \times 10^9\,\text{N m}^2/\text{C}^2$

We need the net force on the charge $-2q$.

2. Find distances from $-2q$ to the other charges

Position of $-2q$ is at $x = \frac{3R}{4}$

So,

• Distance between $q$ at $0$ and $-2q$ at $\frac{3R}{4}$: $r_1 = \frac{3R}{4}$

• Distance between $2q$ at $R$ and $-2q$ at $\frac{3R}{4}$: $r_2 = R - \frac{3R}{4} = \frac{R}{4}$

Since $R=2\times 10^{-2}\,\text{m}$, $r_1 = \frac{3}{4}(2\times10^{-2}) = 1.5\times10^{-2}\,\text{m}$ $r_2 = \frac{1}{4}(2\times10^{-2}) = 0.5\times10^{-2} = 5\times10^{-3}\,\text{m}$

3. Force on $-2q$ due to charge $q$ at the origin

Magnitude: $F_1 = k\frac{|q(-2q)|}{r_1^2} = k\frac{2q^2}{r_1^2}$

Now, $q^2 = (2\times10^{-6})^2 = 4\times10^{-12}$

Thus, $F_1 = 9\times10^9 \cdot \frac{2(4\times10^{-12})}{(1.5\times10^{-2})^2}$ $F_1 = 9\times10^9 \cdot \frac{8\times10^{-12}}{2.25\times10^{-4}}$ $F_1 = \frac{72\times10^{-3}}{2.25\times10^{-4}} = 320\,\text{N}$

Direction: $q$ and $-2q$ attract each other, so the force is towards the origin, i.e. to the left.

4. Force on $-2q$ due to charge $2q$ at $x=R$

Magnitude: $F_2 = k\frac{|(-2q)(2q)|}{r_2^2} = k\frac{4q^2}{r_2^2}$

So, $F_2 = 9\times10^9 \cdot \frac{4(4\times10^{-12})}{(5\times10^{-3})^2}$ $F_2 = 9\times10^9 \cdot \frac{16\times10^{-12}}{25\times10^{-6}}$ $F_2 = \frac{144\times10^{-3}}{25\times10^{-6}} = 5760\,\text{N}$

Direction: $-2q$ and $2q$ attract each other, so the force is towards $x=R$, i.e. to the right.

5. Net force on $-2q$

The two forces are opposite in direction:

• $F_1 = 320\,\text{N}$ left
• $F_2 = 5760\,\text{N}$ right

Hence magnitude of net force is $F_{\text{net}} = 5760 - 320 = 5440\,\text{N}$

6. Final answer

$\boxed{5440}$

So, the required integer answer is 5440.