(a) The electric field strength at a point 1.00 cm to the left of the middle is 2.0 x 10⁷ N/C.
(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.
Electric field strength
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
Net electric field
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
Force on the charge −4.72 µC
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 left.
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