At any point on the right bisector of line joining two equal and opposite charges

Options

A. the electric field is zero
B. the electric potential is zero
C. the electric potential decreases with increasing distance from centre
D. the electric field is perpendicular to the line joining the charges

Answer

To solve the problem regarding the electric field and potential at any point on the right bisector of the line joining two equal and opposite charges, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Let’s denote the two charges as \( +q \) and \( -q \).
- Assume \( +q \) is located at point \( A \) and \( -q \) is located at point \( B \). The distance between them is \( 2a \).

2. **Determine the Right Bisector**:
- The right bisector of the line joining the two charges is a vertical line that bisects the line segment \( AB \) at its midpoint.
- The midpoint is at a distance \( a \) from both charges.

3. **Choose a Point on the Right Bisector**:
- Let’s consider a point \( P \) on the right bisector at a distance \( d \) above the midpoint.

4. **Calculate the Distance from Each Charge to Point P**:
- The distance from point \( P \) to each charge can be calculated using the Pythagorean theorem.
- The distance \( r \) from point \( P \) to either charge is given by:
\[ r = \sqrt{a^2 + d^2} \]

5. **Calculate the Electric Field Due to Each Charge**:
- The electric field \( E \) due to charge \( +q \) at point \( P \) is directed away from the charge:
\[ E_{+q} = \frac{kq}{r^2} = \frac{kq}{(a^2 + d^2)} \]
- The electric field \( E \) due to charge \( -q \) at point \( P \) is directed towards the charge:
\[ E_{-q} = \frac{kq}{r^2} = \frac{kq}{(a^2 + d^2)} \]

6. **Determine the Components of the Electric Fields**:
- Since the configuration is symmetric, the horizontal components of the electric fields from both charges will cancel each other out.
- The vertical components will add up.

7. **Calculate the Net Electric Field**:
- The net electric field \( E_{net} \) at point \( P \) can be expressed as:
\[ E_{net} = E_{+q} + E_{-q} = 2E \cos(\theta) \]
- However, since the horizontal components cancel out, the net electric field at point \( P \) will be zero.

8. **Calculate the Electric Potential**:
- The electric potential \( V \) at point \( P \) due to both charges is given by:
\[ V = V_{+q} + V_{-q} = \frac{kq}{r} - \frac{kq}{r} = 0 \]
- Thus, the electric potential at any point on the right bisector is zero.

### Conclusion:
- At any point on the right bisector of the line joining two equal and opposite charges, the electric field is **zero** and the electric potential is also **zero**.