Identical point charges of `+1.7 mu C` are fixed to diagonally opposite corners of a square. A third charge is then fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without changing magnitudes. Find the sign and magnitude of the third charge.

Options

A. `-4.8 mu C`
B. `-3.4 mu C`
C. `+5.9 mu C`
D. `+1.7 mu C`

Answer

To solve the problem, we need to determine the sign and magnitude of the third charge placed at the center of a square, given that two identical point charges of \( +1.7 \, \mu C \) are fixed at diagonally opposite corners. The potential at the empty corners must change signs without changing magnitudes.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a square with sides of length \( a \).
- Charges \( +1.7 \, \mu C \) are placed at corners \( B \) and \( D \).
- We need to find a charge \( q' \) placed at the center of the square that changes the potential at the empty corners \( A \) and \( C \).

2. **Calculate the Potential at Corner A Without the Center Charge**:
- The potential \( V_A \) at corner \( A \) due to charges at corners \( B \) and \( D \) is given by:
\[ V_A = k \left( \frac{q_1}{r_{AB}} + \frac{q_2}{r_{AD}} \right) \]
- Here, \( q_1 = q_2 = 1.7 \, \mu C \) and the distances \( r_{AB} = a \) and \( r_{AD} = a \).
- Thus,
\[ V_A = k \left( \frac{1.7 \times 10^{-6}}{a} + \frac{1.7 \times 10^{-6}}{a} \right) = 2k \frac{1.7 \times 10^{-6}}{a} = \frac{3.4k}{a} \]

3. **Calculate the Potential at Corner A With the Center Charge**:
- When the charge \( q' \) is placed at the center, the potential at corner \( A \) becomes:
\[ V'_A = k \left( \frac{1.7 \times 10^{-6}}{a} + \frac{1.7 \times 10^{-6}}{a} + \frac{q'}{d} \right) \]
- The distance \( d \) from the center to any corner is \( \frac{a}{\sqrt{2}} \).
- Therefore,
\[ V'_A = k \left( \frac{1.7 \times 10^{-6}}{a} + \frac{1.7 \times 10^{-6}}{a} + \frac{q'}{\frac{a}{\sqrt{2}}} \right) = k \left( \frac{3.4 \times 10^{-6}}{a} + \frac{\sqrt{2}q'}{a} \right) \]

4. **Setting the Condition for Changing Signs**:
- According to the problem, the potential at corner \( A \) after placing the charge should equal the negative of the potential before placing the charge:
\[ V'_A = -V_A \]
- Thus,
\[ k \left( \frac{3.4 \times 10^{-6}}{a} + \frac{\sqrt{2}q'}{a} \right) = -\frac{3.4k}{a} \]

5. **Solving for \( q' \)**:
- Canceling \( k/a \) from both sides (assuming \( k \) and \( a \) are non-zero):
\[ 3.4 \times 10^{-6} + \sqrt{2}q' = -3.4 \times 10^{-6} \]
- Rearranging gives:
\[ \sqrt{2}q' = -3.4 \times 10^{-6} - 3.4 \times 10^{-6} = -6.8 \times 10^{-6} \]
- Therefore,
\[ q' = \frac{-6.8 \times 10^{-6}}{\sqrt{2}} \approx -4.8 \times 10^{-6} \, C \]

6. **Conclusion**:
- The charge \( q' \) is approximately \( -4.8 \, \mu C \).
- Thus, the sign of the third charge is negative, and its magnitude is \( 4.8 \, \mu C \).

### Final Answer:
- **Sign**: Negative
- **Magnitude**: \( 4.8 \, \mu C \)