Question 12 Three positive point charges, A, B and C, one of $4 \mu \mathrm{C}$ and two of $2 \mu \mathrm{C}$ are equally spaced, $120^{\circ}$ apart, on the circumference of a circle of radius $4.33 \times 10^{-2} \mathrm{~m}$. Assume the system of charges is in a vacuum. (B) (C) a) i. Determine the magnitude of the electric field strength at the centre [ 3 ] of the circle due to the $4 \mu \mathrm{C}$ charge. ii. Determine the magnitude of the electric field strength at the centre [ 3 ] of the circle due to either $E_{B}$ or $E_{C}$. iii. Determine the resultant electric field strength at the centre of the [ 4 ] circle, including its direction, due to all three charges.

Step 1 ：Given that the charge of A is \(4 \mu C\), the charge of B and C is \(2 \mu C\), and the radius of the circle is \(4.33 \times 10^{-2} m\).

Step 2 ：Using the formula for electric field strength \(E = k \times \frac{q}{r^2}\), where \(k\) is Coulomb's constant \(8.99 \times 10^9 N m^2/C^2\), \(q\) is the charge, and \(r\) is the distance from the charge to the point.

Step 3 ：For part i, we calculate the electric field strength at the center of the circle due to the 4 μC charge. Substituting the given values into the formula, we get \(E_A = k \times \frac{q_A}{r^2} = 8.99 \times 10^9 \times \frac{4 \times 10^{-6}}{(4.33 \times 10^{-2})^2} = 1.918 \times 10^{7} N/C\).

Step 4 ：For part ii, we calculate the electric field strength at the center of the circle due to either of the 2 μC charges. Substituting the given values into the formula, we get \(E_B = E_C = k \times \frac{q_B}{r^2} = 8.99 \times 10^9 \times \frac{2 \times 10^{-6}}{(4.33 \times 10^{-2})^2} = 9.590 \times 10^{6} N/C\).

Step 5 ：For part iii, since the charges are equally spaced around the circle, the electric fields due to the two 2 μC charges will cancel each other out. So, the resultant electric field will be the same as the electric field due to the 4 μC charge, which is \(E_{\text{resultant}} = E_A = 1.918 \times 10^{7} N/C\).

Step 6 ：Final Answer: \(E_{A} = \boxed{1.918 \times 10^{7} \mathrm{N/C}}\), \(E_{B} = E_{C} = \boxed{9.590 \times 10^{6} \mathrm{N/C}}\), and \(E_{\text{resultant}} = \boxed{1.918 \times 10^{7} \mathrm{N/C}}\).