A charge of 10.104 nC is uniformly distributed along the x-axis from −1 m to 1 m.

What is the electric potential (relative to zero at infinity) of the point at 5 m on the x-axis?

To find the electric potential at a point along the x-axis due to a uniformly distributed charge, we can use the equation for the electric potential due to a continuous charge distribution:

V = k ∫ (λ / r) dx

Where:
V is the electric potential
k is Coulomb's constant (k = 8.99 × 10^9 N m^2/C^2)
λ is the linear charge density (charge per unit length)
r is the distance from the charge element (dx, in this case) to the point at which we want to calculate the electric potential

In this case, the linear charge density λ is equal to the total charge Q divided by the total length L:

λ = Q / L

Given that the charge is uniformly distributed from -1 m to 1 m, the length L is equal to 2 m. The total charge Q is given as 10.104 nC (nanoCoulombs), which is equal to 10.104 × 10^(-9) C.

Substituting these values into the equation, we have:

λ = (10.104 × 10^(-9) C) / (2 m)
λ = 5.052 × 10^(-9) C/m

Now we can calculate the electric potential at the point 5 m on the x-axis:

V = k ∫ (λ / r) dx

Since the charge is distributed along the x-axis from -1 m to 1 m, the integral limits will be -1 m to 1 m.

V = k ∫ (5.052 × 10^(-9) C/m / r) dx, from x = -1 m to 1 m

To find the electric potential at x = 5 m, we need to calculate the integral using these limits. Since the point is outside the region where the charge is distributed, the value of the potential will be the same as the one at x = 1 m, which we need to calculate.

Finally, we can plug in the values into the integral and solve for the electric potential.