contestada

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.

Respuesta :

aguilerapablo03

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