Question

Three point charges of magnitude 4µC, 0.12µC and 0.12µC are located at the vertices A, B, C of right angled triangle whose sides are $AB = \sqrt{3}$ cm, $BC = \sqrt{7}$ cm and $CA = \sqrt{4}$ cm and point A is the right angle corner. Charge at point A experiences _______ N of electrostatic force due to the other two charges.

Answer 1

18

Answer 2

The problem asks to calculate the net electrostatic force experienced by the charge located at vertex A of a right-angled triangle due to the other two charges at B and C.

1. Identify Given Quantities:

• Charge at A, $q_A = 4 \, \mu C = 4 \times 10^{-6} \, C$
• Charge at B, $q_B = 0.12 \, \mu C = 0.12 \times 10^{-6} \, C$
• Charge at C, $q_C = 0.12 \, \mu C = 0.12 \times 10^{-6} \, C$
• Side lengths: $AB = \sqrt{3} \, \text{cm} = \sqrt{3} \times 10^{-2} \, \text{m}$
• $CA = \sqrt{4} \, \text{cm} = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}$
• $BC = \sqrt{7} \, \text{cm} = \sqrt{7} \times 10^{-2} \, \text{m}$
• Point A is the right-angle corner, meaning $\angle A = 90^\circ$. We can verify this using Pythagoras theorem: $AB^2 + CA^2 = (\sqrt{3})^2 + (\sqrt{4})^2 = 3 + 4 = 7$. And $BC^2 = (\sqrt{7})^2 = 7$. Since $AB^2 + CA^2 = BC^2$, the triangle is indeed a right-angled triangle at A.
• Coulomb's constant, $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$.

2. Calculate the Force $F_{AB}$ (Force on $q_A$ due to $q_B$): The distance between A and B is $r_{AB} = \sqrt{3} \times 10^{-2} \, \text{m}$. Using Coulomb's Law, $F = k \frac{|q_1 q_2|}{r^2}$: $F_{AB} = k \frac{|q_A q_B|}{r_{AB}^2}$ $F_{AB} = (9 \times 10^9 \, \text{N m}^2/\text{C}^2) \frac{(4 \times 10^{-6} \, C)(0.12 \times 10^{-6} \, C)}{(\sqrt{3} \times 10^{-2} \, \text{m})^2}$ $F_{AB} = (9 \times 10^9) \frac{0.48 \times 10^{-12}}{3 \times 10^{-4}}$ $F_{AB} = (9 \times 0.16) \times 10^{9 - 12 + 4}$ $F_{AB} = 1.44 \times 10^1 = 14.4 \, \text{N}$ Since both charges are positive, the force $F_{AB}$ is repulsive and directed along the line AB, away from B.

3. Calculate the Force $F_{AC}$ (Force on $q_A$ due to $q_C$): The distance between A and C is $r_{AC} = 2 \times 10^{-2} \, \text{m}$. $F_{AC} = k \frac{|q_A q_C|}{r_{AC}^2}$ $F_{AC} = (9 \times 10^9 \, \text{N m}^2/\text{C}^2) \frac{(4 \times 10^{-6} \, C)(0.12 \times 10^{-6} \, C)}{(2 \times 10^{-2} \, \text{m})^2}$ $F_{AC} = (9 \times 10^9) \frac{0.48 \times 10^{-12}}{4 \times 10^{-4}}$ $F_{AC} = (9 \times 0.12) \times 10^{9 - 12 + 4}$ $F_{AC} = 1.08 \times 10^1 = 10.8 \, \text{N}$ Since both charges are positive, the force $F_{AC}$ is repulsive and directed along the line AC, away from C.

4. Calculate the Net Force $F_{net}$: Since point A is the right-angle corner, the forces $F_{AB}$ and $F_{AC}$ are perpendicular to each other. The net force is the vector sum of these two forces. $F_{net} = \sqrt{F_{AB}^2 + F_{AC}^2}$ $F_{net} = \sqrt{(14.4 \, \text{N})^2 + (10.8 \, \text{N})^2}$ $F_{net} = \sqrt{207.36 + 116.64}$ $F_{net} = \sqrt{324}$ $F_{net} = 18 \, \text{N}$

The charge at point A experiences 18 N of electrostatic force due to the other two charges.

The final answer is $\boxed{\text{18}}$.

Explanation of the solution:

1. Calculate the electrostatic force $F_{AB}$ exerted by charge $q_B$ on $q_A$ using Coulomb's Law: $F_{AB} = k \frac{q_A q_B}{r_{AB}^2}$.
2. Calculate the electrostatic force $F_{AC}$ exerted by charge $q_C$ on $q_A$ using Coulomb's Law: $F_{AC} = k \frac{q_A q_C}{r_{AC}^2}$.
3. Since A is the right-angle corner, the forces $F_{AB}$ and $F_{AC}$ are perpendicular. The net force $F_{net}$ is found using the Pythagorean theorem: $F_{net} = \sqrt{F_{AB}^2 + F_{AC}^2}$.

Answer: 18 N