What is the work done in moving a charge of +2μC from a point A to B in a unifor

₹0.0

Question: What is the work done in moving a charge of +2μC from a point A to B in a uniform electric field of 500 N/C over a distance of 0.4m?

Options:

Correct Answer: 80 J

Solution:

Work done W = F * d = q * E * d = (2 × 10^-6 C) * (500 N/C) * (0.4 m) = 0.4 J = 80 J.

What is the work done in moving a charge of +2μC from a point A to B in a uniform electric field of 500 N/C over a distance of 0.4m?

What is the work done in moving a charge of +2μC from a point A to B in a uniform electric field of 500 N/C over a distance of 0.4m?

Step 1: Identify the charge being moved. In this case, it is +2μC (microcoulombs). Convert this to coulombs: +2μC = 2 × 10^-6 C.
Step 2: Identify the strength of the electric field. Here, it is given as 500 N/C (newtons per coulomb).
Step 3: Identify the distance over which the charge is moved. This distance is 0.4 m (meters).
Step 4: Use the formula for work done in an electric field: Work done (W) = charge (q) × electric field (E) × distance (d).
Step 5: Substitute the values into the formula: W = (2 × 10^-6 C) × (500 N/C) × (0.4 m).
Step 6: Calculate the work done: W = 2 × 10^-6 × 500 × 0.4 = 0.4 J (joules).
Step 7: Note that the final answer is 0.4 J, not 80 J. The correct work done is 0.4 J.

No concepts available.