Numerical: Charge and Coulomb’s Law - Shortcut methods and tricks

JEE Mains - Level

1. Two charges, +2 µC and -4 µC, are separated by a distance of 10 cm in air. Calculate the magnitude of the electrostatic force between them. Shortcut: Use the formula |Kq1q2|/r^2, where K is Coulomb’s constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges in coulombs, and r is the distance between the charges in meters. Solution: Plug in the values: (9 x 10^9 N m^2/C^2)(2 x 10^-6 C)(-4 x 10^-6 C) / (0.1 m)^2 Simplify: (9 x 10^9 N m^2/C^2)(8 x 10^-12 C^2) / (0.01 m^2) Result: 72 x 10^-3 N = 72 mN Therefore, the magnitude of the electrostatic force between the charges is 72 mN.

2. Two identical point charges, each of charge 5 x 10^-6 C, are separated by a distance of 0.1 m in a vacuum. What is the magnitude of the electric field due to one charge at the location of the other? Shortcut: Directly use the formula E = Kq/r^2, where E is the electric field in volts per meter (V/m), K is Coulomb’s constant (9 x 10^9 N m^2/C^2), q is the charge in coulombs, and r is the distance between the charges in meters. Solution: Plug in the values: (9 x 10^9 N m^2/C^2)(5 x 10^-6 C) / (0.1 m)^2 Simplify: (9 x 10^9 N m^2/C^2)(25 x 10^-12 C) / (0.01 m^2) Result: 2.25 x 10^6 V/m Therefore, the magnitude of the electric field due to one charge at the location of the other is 2.25 MV/m.

3. Three charges, +2 µC, +3 µC, and -5 µC, are placed at the vertices of an equilateral triangle of side 15 cm. Calculate the magnitude of the net electrostatic force on the charge at the vertex. Shortcut: Use the vector addition method to calculate the net force. Since the triangle is equilateral, the forces due to the other two charges will be equal in magnitude but opposite in direction. Hence, we can subtract the forces to find the net force. Solution: Calculate the distance between the charges: a = side of the equilateral triangle = 15 cm = 0.15 m Calculate the magnitude of the forces: F = Kq1q2/r^2 = (9 x 10^9 N m^2/C^2)(2 x 10^-6 C)(3 x 10^-6 C) / (0.15 m)^2 Simplify: (9 x 10^9 N m^2/C^2)(6 x 10^-12 C^2) / (0.0225 m^2) Result: F = 2.4 x 10^-3 N Since the forces are in opposite directions, we subtract them: Net force = 2(2.4 x 10^-3 N) - (5 x 10^-6 C)(9 x 10^9 N m^2/C^2) / (0.15 m)^2 Simplify: Net force = 4.8 x 10^-3 N - (45 x 10^-3 N) Net force = -3.72 x 10^-3 N Since the net force is negative, it acts in the opposite direction of the assumed positive force. Therefore, the magnitude of the net electrostatic force on the charge at the vertex is 3.72 mN and acts opposite to the assumed direction.

CBSE Board Exam - Level

4. A point charge of 5 x 10^-9 C is placed in a vacuum. Calculate the magnitude of the electric field at a distance of 10 cm from the charge. Shortcut: Directly use the formula E = Kq/r^2, where E is the electric field in volts per meter (V/m), K is Coulomb’s constant (9 x 10^9 N m^2/C^2), q is the charge in coulombs, and r is the distance between the charges in meters. Solution: Plug in the values: (9 x 10^9 N m^2/C^2)(5 x 10^-9 C) / (0.1 m)^2 Simplify: (9 x 10^9 N m^2/C^2)(25 x 10^-18 C) / (0.01 m^2) Result: 4.5 x 10^2 V/m Therefore, the magnitude of the electric field at a distance of 10 cm from the charge is 450 V/m.

5. Two charges, +4 µC and -2 µC, are separated by a distance of 20 cm in air. What is the magnitude of the electric potential at a point midway between the charges? Shortcut: Use the formula V = Kq/r, where V is the electric potential in volts (V), K is Coulomb’s constant (9 x 10^9 N m^2/C^2), q is the charge in coulombs, and r is the distance between the charge and the point in meters. Solution: Calculate the distance to the midpoint: d = 20 cm / 2 = 10 cm = 0.1 m Calculate the potential due to each charge: V1 = Kq1/d and V2 = Kq2/d Plug in the values: V1 = (9 x 10^9 N m^2/C^2)(4 x 10^-6 C) / 0.1 m = 36 x 10^3 V V2 = (9 x 10^9 N m^2/C^2)(-2 x 10^-6 C) / 0.1 m = -18 x 10^3 V Calculate the net potential: Vnet = V1 + V2 Vnet = 36 x 10^3 V - 18 x 10^3 V = 18 x 10^3 V Therefore, the magnitude of the electric potential at a point midway between the charges is 18 kV.

6. A positive charge of 3 µC is placed in a uniform electric field of strength 500 N/C. Calculate the magnitude of the electric force experienced by the charge. Shortcut: Use the formula F = qE, where F is the electric force in newtons (N), q is the charge in coulombs, and E is the electric field strength in volts per meter (V/m). Solution: Plug in the values: F = (3 x 10^-6 C)(500 N/C) Simplify: F = 1.5 x 10^-3 N Therefore, the magnitude of the electric force experienced by the charge is 1.5 mN.