**When an electron moves from A to B along an electric field line, the electric field does 3.94\times10^{-9}J of work on it. What are the electric potential differences (a) V_{B}-V_{A} (b) V_{C}-V_{A} and (c) V_{C}-V_{B}?**

# Solution:

### Note that equipotential surfaces will always be perpendicular to the electric field lines. The figure above shows the electric field lines and the equipotential surfaces for a uniform electric field.

### a) V_{B} - V_{A} = \triangle \frac{U}{q} = \frac{-W}{-e} = \frac{-(3.94 \times 10^{-19} J)}{(-1.60\times10^{-19} C)} = 2.46 V

### b)Note that V_{C} - V_{A} = V_{B} - V_{A} = 2.46V

### c)V_{C} - V_{B} = 0 because B and C are on the same equipotential line.

why there’s a negative sign with e?

The charge of an electron is negative, (-1e)

How q becomes equal to e?

We are looking at the charge, and since we are looking at electrons, the charge of it is what we use 🙂