A charge of +5μC is placed at the origin. What is the electric field at a point 2m away along the x-axis?
Practice Questions
1 question
Q1
A charge of +5μC is placed at the origin. What is the electric field at a point 2m away along the x-axis?
0 N/C
1125 N/C
2250 N/C
4500 N/C
E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m)² = 1125 N/C.
Questions & Step-by-step Solutions
1 item
Q
Q: A charge of +5μC is placed at the origin. What is the electric field at a point 2m away along the x-axis?
Solution: E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m)² = 1125 N/C.
Steps: 9
Step 1: Identify the charge (q) and its value. Here, q = +5μC, which is equal to 5 × 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| / r², 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 N m²/C²) * (5 × 10^-6 C) / (2 m)².
Step 5: Calculate the denominator: (2 m)² = 4 m².
Step 6: Now calculate the electric field: E = (9 × 10^9) * (5 × 10^-6) / 4.
Step 7: Perform the multiplication: (9 × 10^9) * (5 × 10^-6) = 45,000 N m²/C.
Step 8: Finally, divide by 4: E = 45,000 N m²/C / 4 = 11,250 N/C.
Step 9: Round the answer to three significant figures: E = 1125 N/C.
