A charge of -2.68 ✕ 10-9 C is at the origin of a coordinate system, and a charge of 7.77 ✕ 10-9 C is on the x-axis at 2.00 m. At what two locations on the x-axis is the electric potential zero? (Hint: One location is between the charges, and the other is to the left of the y-axis.)

To find the two locations on the x-axis where the electric potential is zero, we can use the principle of superposition, which states that the electric potential due to multiple charges is the algebraic sum of the electric potentials due to each individual charge.

Let's first calculate the electric potential due to the charge at the origin. The electric potential (V) at a point due to a point charge can be calculated using the formula:

V = k * (q / r)

where k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point.

Given that the charge at the origin is -2.68 x 10^-9 C and the distance is 0 (since it's at the origin), we can calculate the electric potential at any point:

V1 = k * (-2.68 x 10^-9 C) / r1

Now, let's calculate the electric potential due to the charge on the x-axis at 2.00 m. Again, using the same formula:

V2 = k * (7.77 x 10^-9 C) / r2

Now, we want to find the locations on the x-axis where the electric potential is zero. This means that V1 + V2 = 0. Substituting the values, we have:

-2.68 x 10^-9 C / r1 + 7.77 x 10^-9 C / r2 = 0

We have two unknowns, r1 and r2. To solve for them, we'll need another equation. The additional information given suggests that one location is between the charges (let's call it x) and the other location is to the left of the y-axis (let's call it y). This means that:

r1 = x
r2 = 2 - y

Substituting these values into the equation, we get:

-2.68 x 10^-9 C / x + 7.77 x 10^-9 C / (2 - y) = 0

Now, we can solve this equation for x and y. Rearranging the equation, we have:

-2.68 x 10^-9 C / x = 7.77 x 10^-9 C / (2 - y)

Cross-multiplying, we get:

-2.68 x 10^-9 C * (2 - y) = 7.77 x 10^-9 C * x

Expanding and rearranging, we have:

-5.36 x 10^-9 C + 2.68 x 10^-9 C * y = 7.77 x 10^-9 C * x

Now, let's consider the two cases separately:

1. For the location between the charges (x):
In this case, y = 0 since the location is between the charges.
Plugging this into the equation, we have:

-5.36 x 10^-9 C = 7.77 x 10^-9 C * x

Solving for x, we get:

x = -5.36 x 10^-9 C / 7.77 x 10^-9 C
≈ -0.6898 m

Therefore, one location where the electric potential is zero is approximately -0.6898 m on the x-axis.

2. For the location to the left of the y-axis (y):
In this case, x = 0 since the location is to the left of the y-axis.
Plugging this into the equation, we have:

-5.36 x 10^-9 C + 2.68 x 10^-9 C * y = 0

Solving for y, we get:

2.68 x 10^-9 C * y = 5.36 x 10^-9 C
y = 5.36 x 10^-9 C / 2.68 x 10^-9 C
≈ 2.0000 m

Therefore, the other location where the electric potential is zero is approximately 2.0000 m on the x-axis.

In summary, the two locations on the x-axis where the electric potential is zero are approximately -0.6898 m and 2.0000 m.