The potential difference between two points, A and B, in an electric field is 2.00 volts. The energy required to move a charge of 8×10^-19 coulomb from point A to point B is

Answer: 1.60 * 10^-18 J

Explanation:

V= W/q

W = Vq = (2.00 V)(8.00 * 10^-19 C) = 1.60 * 10^-18 J

Let A and B be two points located in a uniform electric field, A being a distance d from B in the direction of the field. The work that an external force must do to bring a unit positive charge q from the reference point to the point considered against the electric force at constant speed, mathematically is expressed by:

[tex]V_B_A=\frac{W_A_B}{q}[/tex]

Therefore, isolating [tex]W_A_B[/tex] and replacing the data provided:

Answer: 1.60 * 10^-18 J

Explanation:

V= W/q

W = Vq = (2.00 V)(8.00 * 10^-19 C) = 1.60 * 10^-18 J

Answer:[tex]W_A_B=-1.6\times 10^{-18} J[/tex]

Explanation:Let A and B be two points located in a uniform electric field, A being a distance d from B in the direction of the field. The work that an external force must do to bring a unit positive charge q from the reference point to the point considered against the electric force at constant speed, mathematically is expressed by:

[tex]V_B_A=\frac{W_A_B}{q}[/tex]

Therefore, isolating [tex]W_A_B[/tex] and replacing the data provided:

[tex]W_A_B=V_B_A *q=-2*(8\times 10^{-19}) =-1.6\times 10^{-18}J[/tex]