Given two particles with Q = 1.90-µC charges as shown in the figure below and a particle with charge q = 1.31 ✕ 10−18 C at the origin. (Note: Assume a reference level of potential V = 0 at r = ∞.)

What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 1.90-µC particles?

To find the change in electric potential energy (ΔU) when moving the charge \( q = 1.31 \times 10^{-18} \, C \) a distance \( d = 0.400 \, m \) closer to either of the two 1.90-µC charges, we can use the concept of electric potential energy in a system of point charges.

### Step 1: Define the electric potential energy formula for point charges.
The electric potential energy \( U \) between two charges \( q_1 \) and \( q_2 \), separated by a distance \( r \), is given by:

\[
U = k_e \frac{q_1 q_2}{r}
\]

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

### Step 2: Calculate the initial potential energy.
Let the initial distance between the charge \( q \) and one of the 1.90-µC charges be \( r_i \), and the final distance after moving closer by 0.400 m be \( r_f = r_i - 0.400 \, m \).

The initial potential energy between \( q \) and one of the 1.90-µC charges is:

\[
U_i = k_e \frac{q \times 1.90 \times 10^{-6}}{r_i}
\]

### Step 3: Calculate the final potential energy.
The final potential energy after moving the charge closer by 0.400 m will be:

\[
U_f = k_e \frac{q \times 1.90 \times 10^{-6}}{r_f}
\]

Where \( r_f = r_i - 0.400 \, m \).

### Step 4: Find the change in potential energy.
The change in electric potential energy \( \Delta U \) is given by the difference between the final and initial energies:

\[
\Delta U = U_f - U_i = k_e q \times 1.90 \times 10^{-6} \left( \frac{1}{r_f} - \frac{1}{r_i} \right)
\]

### Step 5: Substitute values.
- \( q = 1.31 \times 10^{-18} \, C \)
- \( 1.90 \times 10^{-6} \, C \) is the charge of each particle.
- \( d = 0.400 \, m \)

Now, we need the initial distance \( r_i \) between the charge \( q \) and one of the 1.90-µC charges. If this is given in the figure, we can substitute it in and calculate the result.

Do you know the initial distance \( r_i \) from the figure?

Given two particles with Q = 1.90-µC charges as shown in the figure below and a particle with charge q = 1.31 × 10⁻¹⁸ C at the origin. (Note: Assume a reference level of potential V = 0 at r = ∞.) What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 1.90-µC particles?

Calculate the change in electric potential energy when a particle with charge q = 1.31 × 10⁻¹⁸ C is moved 0.400 m closer to a 1.90-µC charge. This solution uses the concept of electric potential energy and Coulomb’s law, providing step-by-step calculations for better understanding. Suitable for undergraduate physics students.

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