Two Charges 2 μC and 4 μC Separated by 30 cm — Find Coulomb Force

Question

Two point charges, $q_1 = 2\,\mu C$ and $q_2 = 4\,\mu C$, are placed in vacuum separated by a distance of $30\,cm$. Find the electrostatic force between them.

Solution — Step by Step

Why This Works

Coulomb’s Law is the electrostatic analogue of Newton’s Law of Gravitation. The force depends on the product of charges (so doubling either charge doubles the force) and falls off as the square of distance (so halving the distance quadruples the force).

The constant $k = 9 \times 10^9\,\text{N m}^2\text{C}^{-2}$ is enormous by everyday standards, which tells you that even microcoulombs of charge create forces in the Newton range. This is why we work with $\mu C$ and $nC$ in practice — assembling a full coulomb of charge in one place is essentially impossible.

Notice we got $\frac{9 \times 10^9 \times 8 \times 10^{-12}}{0.09}$. Writing $0.09 = 9 \times 10^{-2}$ makes the cancellation clean. This bookkeeping habit saves time and prevents arithmetic errors in a 3-hour exam.

Alternative Method

We can also use $k = \frac{1}{4\pi\epsilon_0}$ with $\epsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$. This gives the same numerical result but is slower to compute. For CBSE numericals, $k = 9 \times 10^9$ is the standard shortcut and is perfectly acceptable.

Common Mistake