ABC is a Right angled triangle in which AB=3cm and BC=4cm and right angle is at B. The three charges 15×10^-6C ,12 ×10^-6,-20×10^-6 are placed respectively at A,B and C. Find the net force acting on charge at B due to the remaining charges?

Question

ABC is a Right angled triangle in which AB=3cm and BC=4cm and right angle is at B. The three charges  15×10^-6C ,12 ×10^-6,-20×10^-6 are placed respectively at A,B and C. Find the net force acting on charge at B due to the remaining charges?

Answer

Given:

ABC is a right angled triangle with AB=3cm and BC=4cm.

Charge at A = 15×10^-6C, charge at B = 12×10^-6C, charge at C = -20×10^-6C.

To find: Net force acting on charge at B due to the remaining charges.

Solution:

Step 1: Calculate the distance between the charges.

Distance between A and B, AB = 3cm.

Distance between B and C, BC = 4cm.

Distance between A and C, AC = √(AB^2 + BC^2) = 5cm.

Step 2: Calculate the force between the charges.

Force between A and B, FAB = k(Q1Q2/r^2) = 9×10^9 × (15×10^-6) × (12×10^-6) / (3×10^-2)^2 = 1.08×10^-3 N.

Force between B and C, FBC = k(Q1Q2/r^2) = 9×10^9 × (12×10^-6) × (20×10^-6) / (4×10^-2)^2 = -1.08×10^-3 N (negative because charges are of opposite sign).

Step 3: Calculate the net force on charge at B.

Since charge at B is at the right angle of the triangle, forces FAB and FBC will act at 90 degrees to each other.

Net force on charge at B, FBnet = √(FAB^2 + FBC^2) = √((1.08×10^-3)^2 + (-1.08×10^-3)^2) = 1.53×10^-3 N.

Answer: The net force acting on the charge at B due to the remaining charges is 1.53×10^-3 N.

Answer

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