An electron is at the origin. (a) calculate the electric potential va at point a, x 5 0.250 cm. (b) calculate the electric potential vb at point b, x 5 0.750 cm. what is the potential difference vb 2 va? (c) would a negatively charged particle placed at point a necessarily go through this same potential difference upon reaching point b ? explain

a)  V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V  c) the sign of the potential change

Explanation:

The electrical potential for a point charge

     V = k q / r

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²

a) potential At point x = 0.250 cm = 0.250 10-2m

    V_a =  -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²

    V_a = -5.7536 10⁺⁷ V

b) point x = 0.750 cm = 0.750 10-2

    Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²

    Vb = -1.92 10⁻⁷ V

potemcial difference

    ΔV = Vb- Va

    V_ba = (-5.7536 + 1.92) 10⁻⁷

    V_ba = -3.83 10⁻⁷ V

c) To know what would happen to a particle, let's use the relationship between the potential and the electric field

     ΔV = E d

The force on the particle is

     F = q₀ E

     F = q₀ ΔV / d

We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed

          V_ba = 0.83 10⁻⁷ V

that would be because of both gravity and friction. so     c