Find force on charge q = 1 μC due to unfirmly charged rod as shown in the figure

From the given figure:
- The total charge on the rod is $Q=2\text{μC}=2\times {10}^{-6}\text{C}$.
- The length of the rod is $L=10\text{cm}=0.1\text{m}$.
- The point charge is $q=1\text{μC}=1\times {10}^{-6}\text{C}$.
- The distance of the point charge $q$ from the nearer end of the rod is $a=2\text{cm}=0.02\text{m}$.

To find the electrostatic force on the charge $q$ due to the uniformly charged rod, we consider a small element of length $dx$ on the rod at a distance $x$ from the charge $q$.
The linear charge density of the rod is:
$\lambda =\frac{Q}{L}$

The charge on the small element $dx$ is:
$dq=\lambda dx=\frac{Q}{L}dx$

According to Coulomb's Law, the force $dF$ on charge $q$ due to this element $dq$ is:
$dF=\frac{k·q·dq}{x^2}=\frac{k·q·Q}{L·x^2}dx$

To find the total force, we integrate this expression over the entire length of the rod, from $x=a$ to $x=a+L$:
$F={\int }_a^{a+L}\frac{k·q·Q}{L·x^2}dx$

$F=\frac{k·q·Q}{L}{\left[-\frac{1}{x}\right]}_a^{a+L}$

$F=\frac{k·q·Q}{L}\left(\frac{1}{a}-\frac{1}{a+L}\right)$

$F=\frac{k·q·Q}{L}\left(\frac{L}{a(a+L)}\right)$

$F=\frac{k·q·Q}{a(a+L)}$

Now, substitute the given values:
- $k=9\times {10}^9{\text{N·m}}^2{\text{/C}}^2$
- $q=1\times {10}^{-6}\text{C}$
- $Q=2\times {10}^{-6}\text{C}$
- $a=0.02\text{m}$
- $a+L=0.02\text{m}+0.1\text{m}=0.12\text{m}$

Substitute these values into the force equation:
$F=\frac{(9\times {10}^9)·(1\times {10}^{-6})·(2\times {10}^{-6})}{0.02·0.12}$

$F=\frac{18\times {10}^{-3}}{0.0024}$

$F=\frac{0.018}{0.0024}=7.5\text{N}$

Thus, the electrostatic force on the charge $q$ is 7.5 N.