Question

A positive charge (q = +6.0 µC) starts from point A in a constant electric field and accelerates to point B. The work done by the electric force is WAB = +2.2 × 10-3 J. Determine the potential difference VB - VA between the two points. Be sure to include the proper algebraic sign.

Answer 1

The potential difference between point A and point B is 366.67 V; point B is at a lower potential than point A as derived from the work done WAB and the charge q.

Step-by-step explanation:

The potential difference (also known as voltage) between two points in an electric field, such as point A to point B (VB - VA), can be calculated using the work done by or against the electric field to move a charge q from point A to point B. The formula for work done by the electric force is W = q(VB - VA), and given that work WAB = +2.2 × 10-3 J and the charge q = +6.0 µC (or +6.0 × 10-6 C), we can rearrange the formula to solve for the potential difference: VB - VA = WAB / q.

The calculation yields VB - VA = +2.2 × 10-3 J / (+6.0 × 10-6 C) which equals +366.67 V. Therefore, the electric potential difference between point A and point B is 366.67 V, with point B being at a lower potential than point A since the charge is positive and the work done is positive, indicating that it has moved in the direction of the electric field, from higher to lower potential.

Answer 2

Vb - Va = -366.7 V.

Vab = Va - Vb, the potential of a with respect to b, is equal to the work done by the electric force when a unit of charge moves from a to b, it is given by:

Vab = Va - Vb = Wab/q,

So, in order to determinate the potential difference Vb - Va we have to multiply by -1 both side of the equation above:

- (Va - Vb) = - (Wab/q)

Resulting

Vb - Va = -(Wab/q)

Given a positive charge q = 6.0μC = 6.0x10⁻⁶C, Wab = 2.2x10⁻³J. Determine Vb - Va.

Vb - Va = - (2.2x10⁻³J/6.0x10⁻⁶C)

Vb - Va = -366.7 J/C = -366.7 V