A plastic rod is placed along the horizontal from x = -2.0 m to x = 2.0 m . A charge Q = 5 nC is spread uniformly along the line. Use this information to answer each of the question on the side

What is the value of the electric potential created by the line of charge at a point of coordinates x = 0.8 m and y =2.0 m? Round off your answer to 1 decimal place but do not write the unit. 

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.

A plastic rod is placed along the horizontal from x = -2.0 m to x = 2.0 m. A charge Q = 5 nC is spread uniformly along the line. What is the value of the electric potential created by the line of charge at a point of coordinates x = 0.8 m and y = 2.0 m? Round off your answer to 1 decimal place but do not write the unit.

Learn how to calculate the electric potential at a point due to a uniformly charged plastic rod. The problem involves key concepts in electromagnetism such as continuous charge distribution, Coulomb's Law, and integral calculus. Suitable for university-level physics students.

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