Question

An electron is moved from point A to point B in a uniform electric field and gains 4.66×10-15 J of electrostatic potential energy. Calculate the magnitude of the electrostatic potential difference between the two points. (in V)

Answer 1

Answer:

2.9125 x 10⁴ V

Explanation:

The electrostatic potential difference (ΔV) between two points, say A and B, is denoted by $V_{B}$ - $V_{A}$ and is the quotient of the change in the electrostatic potential energy (Δ$P_{E}$) of a charge Q moved from A to B, and the charge. i.e

ΔV = $V_{B}$ - $V_{A}$ = Δ$P_{E}$ / Q               ------------(i)

From the question;

Δ$P_{E}$ = 4.66 x 10⁻¹⁵J

Q = charge of an electron = -1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

$V_{B}$ - $V_{A}$ = (4.66 x 10⁻¹⁵) / (-1.6 x 10⁻¹⁹)

$V_{B}$ - $V_{A}$ = -2.9125 x 10⁴ V

|$V_{B}$ - $V_{A}$| = 2.9125 x 10⁴ V

Therefore, the magnitude of the electrostatic potential difference between these two points is 2.9125 x 10⁴ V