To find the electric potential $V$ at a point due to a uniformly charged line, we use the formula for the potential created by a continuous charge distribution:

$V = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{r}$

Here:

• $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2$ (Coulomb constant),
• $dq$ is the infinitesimal charge element along the line,
• $r$ is the distance between the charge element and the observation point.

Step 1: Define the charge distribution

• The charge $Q = 5 \, \text{nC} = 5 \times 10^{-9} \, \text{C}$ is uniformly distributed along the rod, which extends from $x = -2.0 \, \text{m}$ to $x = 2.0 \, \text{m}$, giving the total length $L = 4.0 \, \text{m}$.
• The linear charge density $\lambda$ is: $\lambda = \frac{Q}{L} = \frac{5 \times 10^{-9}}{4} \, \text{C/m} = 1.25 \times 10^{-9} \, \text{C/m}$

Step 2: Set up the integral for the potential

• Let the position of the charge element on the rod be at $x'$, and the coordinates of the observation point be $(x = 0.8, y = 2.0)$.
• The distance $r$ from a charge element at $x'$ to the observation point is: $r = \sqrt{(x - x')^2 + y^2} = \sqrt{(0.8 - x')^2 + (2.0)^2}$
• The infinitesimal charge element $dq = \lambda dx'$.

Thus, the potential at the observation point is given by: $V = 9 \times 10^9 \int_{-2.0}^{2.0} \frac{\lambda \, dx'}{\sqrt{(0.8 - x')^2 + 4.0}}$

Step 3: Perform the calculation

Now, I'll evaluate the integral numerically to find the value of the potential.

The electric potential created by the line of charge at the point with coordinates $x = 0.8 \, \text{m}$ and $y = 2.0 \, \text{m}$ is approximately 19.2 (in volts).

Would you like further details or have any questions about this result?

Here are five related questions for further exploration:

1. How does the distance from the rod affect the electric potential at different points?
2. What would the potential be at a point along the x-axis, say at $x = 0$ and $y = 0$?
3. How would the electric field differ from the potential at the same observation point?
4. What is the potential if the charge distribution is non-uniform along the rod?
5. How would this solution change if the rod was positioned vertically instead of horizontally?

Tip: The electric potential at any point is scalar, meaning it doesn't have a direction, unlike the electric field, which is a vector.