Four charges are placed on the corners of a rectangle. What is the resultant force on the positive charge (a = 1.1 m, b = 0.9 m, q = 2.3 × 10-9C)?

F = 93.49 × 10^(-9) N

Explanation:

We are given;

Length of rectangle; a = 1.1 m

Width of rectangle; b = 0.9 m

Charge on rectangle; q = 2.3 × 10^(-9) C

Formula for force is;

F = kq²/r²

Where;

k is a constant = 8.99 x 10^(9) N.m²/C²

a) on the width side, we have;

F = [8.99 x 10^(9) × (2.3 × 10^(-9))²]/0.9²

F = 58.79 × 10^(-9) N

This is on the y-axis

b) on the length side, we have;

F = [8.99 x 10^(9) × (2.3 × 10^(-9))²]/1.1²

F = 39.3 × 10^(-9) N

This is on the x-axis

C) using pythagoras theorem, for the diagonal side, we have c² = a² + b².

Thus;

F = [8.99 x 10^(9) × (2.3 × 10^(-9))²]/(1.1² + 0.9²)

F = 23.54 × 10^(-9) N

Using trigonometric ratios, we can find the angle θ.

tan θ = b/a

tan θ = 0.9/1.1

tan θ = 0.8182

θ = tan^(-1) 0.8182

θ = 39.29°

Resolving along x and y axis, we have;

F_x = 23.54 × 10^(-9) × cos 39.29°

F_x = 18.22 × 10^(-9) N

F_y = 23.54 × 10^(-9) × sin 39.29°

F_y = 14.91 × 10^(-9) N

Resultant force will be;

F = √((Σf_x)² + (ΣF_y)²)

F = √[(18.22 × 10^(-9)) + (39.3 × 10^(-9))]² + [(14.91 × 10^(-9)) + (58.79 × 10^(-9))²]

F = √[(33.0855 × 10^(-16)) + (54.317 × 10^(-16))]

F = 93.49 × 10^(-9) N

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