Question

8 Three point like charges are placed at the corners of a rectangle as shown in the figure, a = 22.0 cm and b = 52.0 cm. Find the magnitude of the electric force exerted on the charge q3. Let qı = +2.70 μ C, q2 =-270 μC, q3 =-470 μC. 0.422 N g1O 42 43

Answer 1

Ans:

Q1 = 2.7 µC

Q2 = -2.7 µC

Q3 = -4.7 µC

The magnitude of force F₁₃ = KQ1Q3/(d)² = 0.5146 N

Hence, the force is directed from charge Q3 towards charge Q1,

The magnitude of force F₂₃ = KQ2Q3/(d)² = 0.5146 N

Hence, the force is directed from charge Q2 towards charge Q3,

Now,

F_23x = 0.5146 N

F_23y = 0 N

Here, F₁₃ subtends an angle

ϴ = tan^-1(a/b) = 67.06⁰

with the -x-axis, and an angle 90 - ϴ with the y-axis. It follows from the conventional laws of vector projection that

F_13x = -F₁₃cos ϴ = -0.2 N

F_13x = F₁₃cos (90 - ϴ) = 0.474 N

The x and y components of the resultant force F3 acting on charge Q3 are given by

F_3x = 0.3146 N

F_3y = 0.474 N

F = ((F_3x)² + (F_3y)²)^1/2

So resultant force is 0.568 N