A point charge `q_1` is held stationary at the origin. A second charge `q_2` is placed at a point a, and the electric potential energy of the pair of charges is `-6.4 xx 10^-8 J`. When the second charge is moved to point b, the electric force on the charge does `4.2 xx 10^-8 J` of work. What is the electric potential energy of the pair of charges when the second charge is at point b?

Explanation

To find the electric potential energy of the pair of charges when the second charge is at point B, we can follow these steps:

### Step 1: Understand the relationship between work done and potential energy
The work done by the electric force when moving a charge from one point to another is related to the change in electric potential energy. The relationship is given by:
\[ W = -\Delta U \]
where \( W \) is the work done, and \( \Delta U \) is the change in potential energy.

### Step 2: Express the change in potential energy
We can express the change in potential energy as:
\[ \Delta U = U_f - U_i \]
where \( U_f \) is the final potential energy at point B, and \( U_i \) is the initial potential energy at point A.

### Step 3: Substitute the known values
From the problem, we know:
- The initial potential energy \( U_i = -6.4 \times 10^{-8} \, J \)
- The work done \( W = 4.2 \times 10^{-8} \, J \)

Using the relationship from Step 1, we can rewrite the equation:
\[ W = U_i - U_f \]
Thus,
\[ U_f = U_i - W \]

### Step 4: Plug in the values
Now substitute the known values into the equation:
\[ U_f = (-6.4 \times 10^{-8}) - (4.2 \times 10^{-8}) \]

### Step 5: Perform the calculation
Calculating the right-hand side:
\[ U_f = -6.4 \times 10^{-8} - 4.2 \times 10^{-8} = -10.6 \times 10^{-8} \, J \]

### Final Answer
The electric potential energy of the pair of charges when the second charge is at point B is:
\[ U_f = -10.6 \times 10^{-8} \, J \]

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