A point charge `q_1=q` is placed at point P. Another point charge `q_2=-q` is placed at point Q. At some point `R(R!=P, R!=Q)`, electric potential due to `q-1 is V_1` and electric potential due to `q_2` is `V_2`. Which of the following is correct

Options

A. only for some points `V_1gtV_2`
B. only fro some points `V_2gtV_1`
C. for all points `V_1gtV_2`
D. for all points `V_2gtV_1`

Explanation

To solve the problem, we need to analyze the electric potentials due to the two point charges at a point R, which is not equal to the locations of the charges P and Q.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- We have two point charges:
- Charge \( q_1 = q \) (positive charge) located at point P.
- Charge \( q_2 = -q \) (negative charge) located at point Q.

2. **Understanding Electric Potential:**
- The electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by the formula:
\[ V = k \frac{Q}{r} \]
where \( k \) is Coulomb's constant.

3. **Calculate the Electric Potential at Point R:**
- Let the distance from charge \( q_1 \) (at P) to point R be \( r_1 \) and the distance from charge \( q_2 \) (at Q) to point R be \( r_2 \).
- The electric potential \( V_1 \) at point R due to charge \( q_1 \) is:
\[ V_1 = k \frac{q}{r_1} \]
- The electric potential \( V_2 \) at point R due to charge \( q_2 \) is:
\[ V_2 = k \frac{-q}{r_2} \]

4. **Comparing the Potentials:**
- Since \( q_1 \) is positive, \( V_1 \) will be a positive value.
- Since \( q_2 \) is negative, \( V_2 \) will be a negative value.
- Therefore, we can conclude:
\[ V_1 > 0 \quad \text{and} \quad V_2 < 0 \]

5. **Conclusion:**
- Since \( V_1 \) (positive) is always greater than \( V_2 \) (negative) for any point R that is not equal to P or Q, we can state that:
\[ V_1 > V_2 \quad \text{for all points R} \]

### Final Answer:
The correct option is that \( V_1 \) is greater than \( V_2 \) for all points \( R \).