Question

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

Answer 1

The electric potential at points A and B can be calculated using the formula VA = k * q / rA and VB = k * q / rB. The potential difference VB - VA is the same at both points. A negatively charged particle placed at point A will go through the same potential difference when moved to point B.

Step-by-step explanation:

The electric potential at point A, which is located at x = 0.250 cm, can be calculated using the formula:

VA = k * q / rA

where k is the electrostatic constant, q is the charge of the electron, and rA is the distance from the origin to point A. Similarly, the electric potential at point B, which is located at x = 0.750 cm, can be calculated as VB = k * q / rB. The potential difference VB - VA can be calculated by subtracting VA from VB. Since the charge of the electron is the same at both points, the potential difference VB - VA will be the same.

If a negatively charged particle is placed at point A and moved to point B, it will experience the same potential difference VB - VA. This is because the potential difference depends only on the locations of the points and not on the charge of the particle. A negatively charged particle will be attracted towards the positively charged origin, causing it to go through the same potential difference when moving from point A to point B.

Answer 2

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

Step-by-step 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