A charge of +4μC is placed at the origin. What is the electric field at a point 2m away on the x-axis?

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Q: A charge of +4μC is placed at the origin. What is the electric field at a point 2m away on the x-axis?

Solution: E = k * |q| / r^2 = (9 × 10^9) * 4 × 10^-6 / (2)^2 = 450 N/C.

Steps: 8

Step 1: Identify the charge (q) and its value. Here, q = +4μC, which is equal to 4 × 10^-6 C.
Step 2: Identify the distance (r) from the charge to the point where we want to find the electric field. Here, r = 2m.
Step 3: Use the formula for electric field (E) due to a point charge: E = k * |q| / r^2, where k is the electrostatic constant (approximately 9 × 10^9 N m²/C²).
Step 4: Substitute the values into the formula: E = (9 × 10^9) * (4 × 10^-6) / (2)^2.
Step 5: Calculate the denominator: (2)^2 = 4.
Step 6: Now calculate the electric field: E = (9 × 10^9) * (4 × 10^-6) / 4.
Step 7: Simplify the calculation: E = (9 × 10^9 * 4 × 10^-6) / 4 = 9 × 10^9 * 10^-6 = 450 N/C.
Step 8: Conclude that the electric field at the point 2m away on the x-axis is 450 N/C.