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To solve this problem, we need to calculate the change in electric potential energy (\(\Delta U\)) when a charge is moved in the electric field created by another charge. The formula for electric potential energy between two point charges is:

\[
U = \frac{k \cdot q_1 \cdot q_2}{r}
\]

where:
- \(k\) is the Coulomb's constant, \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\),
- \(q_1\) and \(q_2\) are the charges,
- \(r\) is the distance between the charges.

### Given Data
- Charge \(q_1 = 17.3 \, \text{nC} = 17.3 \times 10^{-9} \, \text{C}\),
- Charge \(q_2 = 10.9 \, \text{C}\),
- Initial distance \(r_1 = 10.9 \, \text{m}\),
- Final distance \(r_2 = 4.2 \, \text{m}\).

### Steps
1. **Calculate Initial Potential Energy** (\(U_1\)) at distance \(r_1\):
   \[
   U_1 = \frac{k \cdot q_1 \cdot q_2}{r_1} = \frac{8.99 \times 10^9 \times 17.3 \times 10^{-9} \times 10.9}{10.9}
   \]

2. **Calculate Final Potential Energy** (\(U_2\)) at distance \(r_2\):
   \[
   U_2 = \frac{k \cdot q_1 \cdot q_2}{r_2} = \frac{8.99 \times 10^9 \times 17.3 \times 10^{-9} \times 10.9}{4.2}
   \]

3. **Calculate the Change in Potential Energy** (\(\Delta U = U_2 - U_1\)).

Let's compute these values.

### Solution

1. **Calculate \(U_1\):**
   \[
   U_1 = \frac{8.99 \times 10^9 \times 17.3 \times 10^{-9} \times 10.9}{10.9} = 1.55427 \, \text{J}
   \]

2. **Calculate \(U_2\):**
   \[
   U_2 = \frac{8.99 \times 10^9 \times 17.3 \times 10^{-9} \times 10.9}{4.2} = 4.0348 \, \text{J}
   \]

3. **Calculate \(\Delta U\):**
   \[
   \Delta U = U_2 - U_1 = 4.0348 - 1.55427 = 2.5 \, \text{J (rounded to 1 decimal place)}
   \]

### Answer
The change in potential energy of the charge is **2.5 J**.

Would you like more details or have any questions?

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### Related Questions

1. How does the electric potential energy change with respect to distance?
2. What would be the potential energy change if the distances were halved?
3. How does the magnitude of charges affect the potential energy between them?
4. What is the significance of Coulomb's constant in these calculations?
5. How does the direction of movement (closer or farther) affect the potential energy?

### Tip
When working with electric potential energy, remember that decreasing the distance between two like charges increases the potential energy, while increasing the distance decreases it.

A 17.3 nC charge is moved from a distance of 10.9 m to 4.2 m away from a 10.9 C charge. What is the change in potential energy of the charge? (please provide your answer to 1 decimal place)

Calculate the change in potential energy when a 17.3 nC charge moves from a distance of 10.9 m to 4.2 m away from a 10.9 C charge. This problem covers electric potential energy and Coulomb's law, suitable for physics students in grades 11-12.

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