Two charges are placed on the x axis. One of the charges (q1 = +9.3 µC) is at x1 = +3.4 cm and the other (q2 = -20 µC) is at x2 = +8.9 cm.

(a) Find the net electric field (magnitude and direction) at x = 0 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
Answer in N/C

(b) Find the net electric field (magnitude and direction) at x = +5.4 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
Answer in N/C

To find the net electric field at a given point, we need to calculate the electric field created by each charge at that point and then vectorially add them together.

The formula for the electric field created by a point charge is:

E = k * (q / r^2)

where E is the electric field, k is the electrostatic constant (9.0 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point where we want to find the electric field.

(a) To find the net electric field at x = 0 cm, we need to calculate the electric field created by each charge at this point.

For q1 (+9.3 µC) at x1 = +3.4 cm:
r1 = |x - x1| = |0 - 3.4| = 3.4 cm = 0.034 m

E1 = k * (q1 / r1^2) = 9.0 x 10^9 Nm^2/C^2 * (9.3 x 10^-6 C / (0.034 m)^2) = 7.38 x 10^3 N/C

Note: The magnitude of the electric field is always positive, and the direction of the electric field is determined by the sign of the charge. Since q1 is positive, the direction of E1 is towards the positive x-axis.

For q2 (-20 µC) at x2 = +8.9 cm:
r2 = |x - x2| = |0 - 8.9| = 8.9 cm = 0.089 m

E2 = k * (q2 / r2^2) = 9.0 x 10^9 Nm^2/C^2 * (-20 x 10^-6 C / (0.089 m)^2) = -4.46 x 10^3 N/C

Note: The magnitude of the electric field is always positive, and the direction of the electric field is determined by the sign of the charge. Since q2 is negative, the direction of E2 is towards the negative x-axis.

To find the net electric field at x = 0 cm, we need to sum the individual electric fields:

E_net = E1 + E2 = (7.38 x 10^3 N/C) + (-4.46 x 10^3 N/C) = 2.92 x 10^3 N/C

Therefore, the net electric field at x = 0 cm is 2.92 x 10^3 N/C in the positive x-axis direction.

(b) To find the net electric field at x = +5.4 cm, we repeat the same process for both charges.

For q1 (+9.3 µC) at x1 = +3.4 cm:
r1 = |x - x1| = |5.4 - 3.4| = 2.0 cm = 0.02 m

E1 = k * (q1 / r1^2) = 9.0 x 10^9 Nm^2/C^2 * (9.3 x 10^-6 C / (0.02 m)^2) = 2.32 x 10^4 N/C

For q2 (-20 µC) at x2 = +8.9 cm:
r2 = |x - x2| = |5.4 - 8.9| = 3.5 cm = 0.035 m

E2 = k * (q2 / r2^2) = 9.0 x 10^9 Nm^2/C^2 * (-20 x 10^-6 C / (0.035 m)^2) = -9.37 x 10^3 N/C

To find the net electric field at x = +5.4 cm, we need to sum the individual electric fields:

E_net = E1 + E2 = (2.32 x 10^4 N/C) + (-9.37 x 10^3 N/C) = 1.39 x 10^4 N/C

Therefore, the net electric field at x = +5.4 cm is 1.39 x 10^4 N/C in the positive x-axis direction.