Two ideantical point charges are placed at a separation of `d`. `P` is a point on the line joining the charges, at a distance `x` from any one charge. The field at `P` is `E, E` is plotted against `x` for value of `x` from close to zero to slightly less then `d`. Which of the following represents the resulting curve

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the electric field \( E \) at a point \( P \) located on the line joining two identical point charges \( +Q \) separated by a distance \( d \). The point \( P \) is at a distance \( x \) from one of the charges.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical point charges \( +Q \) placed at a distance \( d \) apart.
- Point \( P \) is located on the line joining the charges at a distance \( x \) from one charge.

2. **Electric Field Due to Point Charges**:
- The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.

3. **Calculating Electric Fields at Point \( P \)**:
- Let \( E_1 \) be the electric field at point \( P \) due to the charge on the left (let's call it \( Q_1 \)) and \( E_2 \) be the electric field at point \( P \) due to the charge on the right (let's call it \( Q_2 \)).
- The distance from \( Q_1 \) to \( P \) is \( x \).
- The distance from \( Q_2 \) to \( P \) is \( d - x \).
- Therefore, we have:
\[ E_1 = \frac{kQ}{x^2} \]
\[ E_2 = \frac{kQ}{(d - x)^2} \]

4. **Net Electric Field at Point \( P \)**:
- The net electric field \( E \) at point \( P \) is given by:
\[ E = E_1 - E_2 = \frac{kQ}{x^2} - \frac{kQ}{(d - x)^2} \]
- This equation shows that the electric field depends on the distance \( x \).

5. **Behavior of the Electric Field**:
- As \( x \) approaches \( 0 \) (very close to \( Q_1 \)), \( E_1 \) becomes very large and \( E_2 \) is small, thus \( E \) is positive.
- At \( x = \frac{d}{2} \) (the midpoint), \( E_1 = E_2 \) and hence \( E = 0 \).
- As \( x \) approaches \( d \) (very close to \( Q_2 \)), \( E_1 \) becomes small and \( E_2 \) becomes large, thus \( E \) becomes negative.

6. **Graphing \( E \) vs \( x \)**:
- The graph of \( E \) vs \( x \) will start from a positive value when \( x \) is close to \( 0 \), decrease to \( 0 \) at \( x = \frac{d}{2} \), and then become negative as \( x \) approaches \( d \).
- This behavior indicates that the graph will have a shape that starts high, crosses the x-axis (where \( E = 0 \)), and then goes low into negative values.

### Conclusion:
The resulting curve representing \( E \) plotted against \( x \) will be a continuous curve that starts from a positive value, crosses zero at the midpoint, and then goes into negative values as \( x \) approaches \( d \). The correct option representing this curve is option C.