The electric field strength at a point 1.00 cm to the left of the middle is -2.0 x 10⁷ N/C.
The magnitude of the force is 94.4 N and direction of the force on it towards the right.
<h3>
Electric field strength</h3>
The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first charge
E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second charge
E2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third charge
E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
<h3>Net electric field</h3>
E = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C
E = -2.0 x 10⁷ N/C
<h3>Force on the charge −4.72 µC</h3>
F = Eq
F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = 94.4 N
Thus, the direction of the force will be towards the right.
Learn more about force on charge here: brainly.com/question/25923373
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