Answer

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**Hint**The force exerted by the charges on each other can be determined by using the force of the attraction between the two charges formula, by using this formula and the given information in the question, the force exerted can be determined.

**Useful formula**

The force exerted by the charges on each other can be given by,

$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

Where, $F$ is the force exerted between the two charges, ${q_1}$ and ${q_2}$ are the two point charges and $r$ is the distance between the two point charges.

**Complete step by step solution**

Given that,

The first charge is placed at the top of your college building, ${q_1} = 1.0\,C$,

The second charge is placed at the top of your house, ${q_2} = 1.0\,C$,

The distance between the two point charges is, $r = 2\,km = 2000\,m$.

Now,

The force exerted by the charges on each other can be given by,

$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}\,................\left( 1 \right)$

In the above equation (1), the term $\dfrac{1}{{4\pi {\varepsilon _0}}}$ is the constant term and the values of this term is $9 \times {10^9}\,N{m^2}{C^{ - 2}}$. Then,

$F = 9 \times {10^9} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$

By substituting the two point charges and the distance between the two point charges in the above equation, then the above equation is written as,

\[F = \dfrac{{9 \times {{10}^9} \times 1 \times 1}}{{{{2000}^2}}}\]

By multiplying the terms in the above equation, then the above equation is written as,

\[F = \dfrac{{9 \times {{10}^9}}}{{{{2000}^2}}}\]

By squaring the terms in the above equation, then the above equation is written as,

\[F = \dfrac{{9 \times {{10}^9}}}{{4 \times {{10}^6}}}\]

By dividing the terms in the above equation, then the above equation is written as,

\[F = 2.25 \times {10^3}\,N\]

**Thus, the above equation shows the force exerted by the two point charges.**

**Note**The force exerted by the two point charges which are separated by the distance is directly proportional to the two charges and inversely proportional to the distance between them. As the charge increases, the force also increases. As the distance between the two charges increases, the force decreases.

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