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In the figure, the charge q=5.36×10−9Cq=5.36×10−9C. Find the electric potential at x=0.350mx=0.350m.

To solve the problem of finding the electric potential at \( x = 0.350 \, \text{m} \), we need to use the formula for the electric potential due to point charges. The general formula for the electric potential \( V \) at a distance \( r \) from a point charge \( q \) is given by:

\[
V = \frac{k \cdot q}{r}
\]

where:
- \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \) is Coulomb's constant,
- \( q \) is the charge, and
- \( r \) is the distance between the point of interest and the charge.

### Step-by-Step Solution:

1. **Identify the charges**: 
   - At \( x = 0 \, \text{m} \), we have a charge of \( -2q \).
   - At \( x = 1.00 \, \text{m} \), we have a charge of \( +5q \).

   The charge value \( q = 5.36 \times 10^{-9} \, \text{C} \).

2. **Determine the distances**:
   - The distance between \( x = 0.350 \, \text{m} \) and the charge \( -2q \) at \( x = 0 \, \text{m} \) is:
     \[
     r_1 = 0.350 \, \text{m}
     \]
   - The distance between \( x = 0.350 \, \text{m} \) and the charge \( +5q \) at \( x = 1.00 \, \text{m} \) is:
     \[
     r_2 = 1.00 - 0.350 = 0.650 \, \text{m}
     \]

3. **Find the electric potentials**:
   - The potential due to the charge \( -2q \) at \( x = 0 \, \text{m} \) is:
     \[
     V_1 = \frac{k \cdot (-2q)}{r_1} = \frac{8.99 \times 10^9 \cdot (-2 \cdot 5.36 \times 10^{-9})}{0.350}
     \]
   - The potential due to the charge \( +5q \) at \( x = 1.00 \, \text{m} \) is:
     \[
     V_2 = \frac{k \cdot (5q)}{r_2} = \frac{8.99 \times 10^9 \cdot (5 \cdot 5.36 \times 10^{-9})}{0.650}
     \]

4. **Add the potentials** to find the total potential at \( x = 0.350 \, \text{m} \):
   \[
   V = V_1 + V_2
   \]

Now, let's compute the values.The total electric potential at \( x = 0.350 \, \text{m} \) is approximately \( 95.31 \, \text{V} \).

Would you like further details or have any other questions?

Here are some related questions to expand your understanding:
1. How does the sign of a charge affect the electric potential it produces?
2. What would the electric potential be at \( x = -0.350 \, \text{m} \)?
3. How does the distance between the charge and the point of interest influence the potential?
4. What is the difference between electric potential and electric field?
5. How would the potential change if both charges were positive?

**Tip:** The electric potential at a point is a scalar quantity, so you can simply add the potentials from multiple charges algebraically.

In the figure, the charge q=5.36×10−9 C. Find the electric potential at x=0.350 m.

Learn how to calculate the electric potential at a point using the principle of superposition and Coulomb's law for point charges. This example considers charges of -2q and +5q positioned at specified distances, with calculations appropriate for college-level physics.

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