Question

Calculate the work done to move 500 × 10^18 electrons between two points in an electric field where the potential difference between the two points is 1 millivolt (e=1.6×10^19 C)​

Answer 1

Answer:

The work done is 0.08 J.

Explanation:

Given:

Number of electrons (n) = 500 × 10¹⁸

Charge of each electron (e) = 1.6 × 10–¹⁹

Potential difference between the two points = 1 millivolt = 10–³ V

To find:

Work done.

Solution:

We already know the charge of each electron and the total number of electrons. So firstly we will find the total charge then we will find the work done.

Total charge (q) = ne

= 500 × 10¹⁸ × 1.6 × 10–¹⁹ = 80 C

Now work done (w) = ΔVq = 10–³ × 80

= 0.08 J