[Grade 12] electric forces?

The symbol for a coulomb is 'C', not 'Cb'.

Suppose the rectangle is:

A. . B

C. . D

with AB = CD = 0.4m and AC = BD = 0.3m.

The diagonal is BC = AD = 0.5m (3-4-5 triangle)

Or do it the long way using Pythagoras:

BC² = 0.4² + 0.3² = 0.25

BC = 0.5m

Note that if we call angle ABC, ‘θ’ then

sinθ = AC/CB = 0.3/0.5 = 0.6

cosθ = AB/CB = 0.4/0.5 = 0.8

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I’ll use exponential notation so, for example, 3x10⁸ is written as 3e8.

Force on qB from qA is k.qA.qB/r² = 8.99e9 * 2.20 * 2.20 /0.4² = 2.719e11 N

This repulsive force on qB acts from A to B, in the +x direction so has:

x component = 2.719e11 N

y component = 0

Force on qB from qD is k.qD.qB/r² = 8.99e9 * 2.20 * 2.20 /0.3² = 4.835e11 N

This repulsive force on qB acts from D to B, in the +y direction so has:

x component = 0

y component = 4.835e11 N

Force on qB from qC is k.qC.qB/r² = 8.99e9 * 2.20 * 2.20 /0.5² = 1.74e11 N

This repulsive force on qB acts from C to B so has:

x component = 1.74e11cosθ = 1.392e11 N

y component = 1.74e11sinθ = 1.044e11 N

Total of x components: X = 2.719e11 + 0 + 1.392e11 = 4.111e11 N

Total of y components: Y = 0 + 4.835e11 + 1.044e11 = 5.879e11 N

Resultant force F = √(X² + Y²)

= √(4.111e11² + 5.879e11²) = 7.17e11 N

The angle α of F (anticlockwise from the x axis) is given by:

tan(α) = Y/X = 5.879e11/4.111e11 = 1.43

α = tan⁻¹(1.43) = 55.0°

Adding vectors using components as above is a basic skill need for many problems. If you are unsure about adding vectors this way, try the video in the link.