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)

2.9125 x 10⁴ V

Explanation:

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

ΔV = - = Δ / Q               ------------(i)

From the question;

Δ = 4.66 x 10⁻¹⁵J

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

Substitute these values into equation (i) as follows;

- = (4.66 x 10⁻¹⁵) / (-1.6 x 10⁻¹⁹)

- = -2.9125 x 10⁴ V

| - | = 2.9125 x 10⁴ V

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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

would be false hope this out