CBSE Class 12 Commerce — Accountancy, Business Studies, Economics

Question Summary

Four point charges are placed at the corners of a square of side $10\,\text{cm}$. A test charge of $1\,\mu\text{C}$ sits at the centre. We must find the net Coulomb force on the central charge due to all four corner charges.

Definition of Variables

• $q_A = q_C = +2\,\mu\text{C} = +2\times 10^{-6}\,\text{C}$
• $q_B = q_D = -5\,\mu\text{C} = -5\times 10^{-6}\,\text{C}$
• Side of square: $a = 10\,\text{cm} = 0.1\,\text{m}$
• Test charge at centre: $q_0 = 1\,\mu\text{C} = 1\times 10^{-6}\,\text{C}$
• Distance from each corner to centre: $r = \dfrac{a\sqrt{2}}{2} = \dfrac{0.1\sqrt{2}}{2}\,\text{m}$
• Coulomb constant: $k = 9\times 10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$

Core Concepts & Problem-Solving Strategy

• Coulomb's law: $F = \dfrac{k\,|q_1 q_2|}{r^2}$ gives the magnitude of force between two point charges.
• Superposition principle: the net force on the central charge is the vector sum of forces due to each corner charge.
• Symmetry: $A$ and $C$ are diagonally opposite with equal charges; $B$ and $D$ are also diagonally opposite with equal charges. Forces from each diagonal pair on the centre are equal in magnitude but opposite in direction, so they cancel.

Step-by-Step Solution

1. Compute the distance from any corner to the centre:
   $$r = \frac{a\sqrt{2}}{2} = \frac{0.1\times \sqrt{2}}{2} \approx 0.0707\,\text{m}$$
2. Force on $q_0$ due to $q_A$ (along $CA$, since $q_A$ is positive and repels $q_0$):
   $$F_A = \frac{k\,q_A\,q_0}{r^2}$$
3. Force on $q_0$ due to $q_C$ has the same magnitude but points along $AC$, directly opposite to $F_A$:
   $$\vec{F}_A + \vec{F}_C = 0$$
4. Similarly, $q_B$ and $q_D$ are equal negative charges on the other diagonal, so they attract $q_0$ with equal and opposite forces:
   $$\vec{F}_B + \vec{F}_D = 0$$
5. Net force on the central charge:
   $$\vec{F}_{\text{net}} = \vec{F}_A + \vec{F}_B + \vec{F}_C + \vec{F}_D = 0$$

Result: The net force on the $1\,\mu\text{C}$ charge at the centre of the square is $\vec{F}_{\text{net}} = 0$.