What is the force between two small charged spheres having charges of 4 x 10^-7C and 6 x 10^-7C placed 60 cm apart in air?

The correct answer is:

$\begin{array}{l}6��{10}^{-3}N\end{array}$

Step-by-Step Explanation:

To find the electrostatic force between two small charged spheres, we use Coulomb's Law. According to Coulomb's Law, the force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law in air or vacuum is given by:
$$F=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{r^2}$$

Let's identify the given values from the problem statement:
- First charge, $q_1=4\times {10}^{-7}\text{C}$
- Second charge, $q_2=6\times {10}^{-7}\text{C}$
- Distance between the spheres, $r=60\text{cm}=0.6\text{m}$
- Coulomb's constant, $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{N}·{\text{m}}^2/{\text{C}}^2$

Now, substitute these values into Coulomb's equation:
$$F=(9\times {10}^9)\times \frac{(4\times {10}^{-7})\times (6\times {10}^{-7})}{{\left(0.6\right)}^2}$$

Simplify the numerator:
$$\text{Numerator}=9\times 4\times 6\times {10}^{9-7-7}=216\times {10}^{-5}$$

Simplify the denominator:
$${\left(0.6\right)}^2=0.36=36\times {10}^{-2}$$

Divide the simplified numerator by the simplified denominator to find the force $F$:
$$F=\frac{216\times {10}^{-5}}{36\times {10}^{-2}}$$

Perform the division:
$$F=6\times {10}^{-5-(-2)}=6\times {10}^{-3}\text{N}$$

Since both charges are positive, the force between them is repulsive in nature, with a magnitude of $6\times {10}^{-3}\text{N}$.