A charge of +10μC is placed in a uniform electric field of strength 500 N/C. Wha
Practice Questions
Q1
A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?
10 J
1 J
100 J
0.5 J
Questions & Step-by-Step Solutions
A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?
Step 1: Identify the charge value, which is +10μC (microcoulombs). Convert this to coulombs: 10μC = 10 × 10^-6 C.
Step 2: Identify the strength of the electric field, which is 500 N/C.
Step 3: Identify the distance the charge is moved, which is 2 meters.
Step 4: Use the formula for work done (W) in an electric field: W = F * d, where F is the force and d is the distance.
Step 5: Calculate the force (F) using the formula F = E * q, where E is the electric field strength and q is the charge. So, F = 500 N/C * 10 × 10^-6 C.
Step 6: Calculate the force: F = 500 * 10 × 10^-6 = 0.005 N.
Step 7: Now, substitute the force (F) and distance (d) into the work done formula: W = F * d = 0.005 N * 2 m.
Step 8: Calculate the work done: W = 0.005 * 2 = 0.01 J.
Electric Field and Work – Understanding how work is calculated in an electric field using the formula W = F * d, where F is the force on the charge and d is the distance moved.
Charge and Electric Field Interaction – Recognizing the relationship between charge, electric field strength, and the force exerted on the charge.
