An electron is at the origin.

(a) Calculate the electric potential VA at point A, x = 0.410 cm.
(b) Calculate the electric potential VB at point B, x = 0.820 cm. What is the potential difference VB - VA?

In the book the examples they give have a charge to them, but they don't offer any help with this question because the don't list a charge of the electron

Your book does indeed surely give the charge on an electron

e = - 1.6 * 10^-19 Coulombs

In the equations listed they are refering to a point charge which is different then that of 1.6*10^-19. If I were to use that in the equation it would not work. The point charge in the example is 5.00*10^-6 C

never mind wrong question for the previous response. Thanks.

To calculate the electric potential at point A and B, we need to use the formula for the electric potential due to a point charge. The electric potential at a distance r from a point charge Q is given by:

V = k * Q / r

Where:
V is the electric potential,
k is the electrostatic constant (9 x 10^9 Nm^2/C^2),
Q is the charge,
r is the distance from the charge.

(a) To calculate the electric potential at point A, which is x = 0.410 cm, we need to know the charge of the electron. The charge of an electron is -1.6 x 10^-19 C.

V_A = k * Q / r_A

Substituting the values:
V_A = (9 x 10^9 Nm^2/C^2) * (-1.6 x 10^-19 C) / (0.410 cm)

Note: It is important to convert cm to meters in order to maintain consistent units.

1 cm = 0.01 m

V_A = (9 x 10^9 Nm^2/C^2) * (-1.6 x 10^-19 C) / (0.410 cm * 0.01 m/cm)

Now, we can calculate V_A using a calculator.

(b) Similarly, to calculate the electric potential at point B, which is x = 0.820 cm, we can use the same formula:

V_B = k * Q / r_B

V_B = (9 x 10^9 Nm^2/C^2) * (-1.6 x 10^-19 C) / (0.820 cm * 0.01 m/cm)

Again, use a calculator to find V_B.

To calculate the potential difference VB - VA, subtract the potential at point A from the potential at point B:

VB - VA = V_B - V_A

Now, substitute the calculated V_B and V_A into the equation to find the potential difference.