The electric potential `V(z, y,z)` for a planar charge distribution is given by:
`V(x, y, z)={(0,,"for "x lt -d),(-V_(0)(1+x/d)^(2),,"for "-d le x le 0),(-V_(0)(1+2x/d),,"for "0 le x lt d),(-3V_(0),,"for " x gt d):}`
where `-V_(0)` is the potential at the origin and d is a distance. Graph of electric field as a function of position is given as

Options

A. `alpha` is independent of k
B. `alpha` is independent of l
C. `alpha` is independent of both k and l
D. None of the above

Answer

To solve the problem, we need to analyze the electric potential \( V(x, y, z) \) given for different ranges of \( x \) and then derive the electric field \( E(x) \) from the potential. The electric field is related to the electric potential by the equation:

\[ E(x) = -\frac{dV}{dx} \]

### Step-by-Step Solution:

1. **Identify the Regions of the Potential Function**:
The potential \( V(x) \) is defined piecewise for different ranges of \( x \):
- For \( x < -d \): \( V(x) = 0 \)
- For \( -d \leq x \leq 0 \): \( V(x) = -V_0 \left(1 + \frac{x}{d}\right)^2 \)
- For \( 0 < x < d \): \( V(x) = -V_0 \left(1 + \frac{2x}{d}\right) \)
- For \( x \geq d \): \( V(x) = -3V_0 \)

2. **Calculate the Electric Field in Each Region**:
- **For \( x < -d \)**:
\[ E(x) = -\frac{dV}{dx} = -\frac{d(0)}{dx} = 0 \]

- **For \( -d \leq x \leq 0 \)**:
\[ V(x) = -V_0 \left(1 + \frac{x}{d}\right)^2 \]
To find \( E(x) \):
\[ E(x) = -\frac{d}{dx} \left[-V_0 \left(1 + \frac{x}{d}\right)^2\right] = V_0 \cdot 2 \left(1 + \frac{x}{d}\right) \cdot \frac{1}{d} \]
Simplifying gives:
\[ E(x) = \frac{2V_0}{d} \left(1 + \frac{x}{d}\right) \]

- **For \( 0 < x < d \)**:
\[ V(x) = -V_0 \left(1 + \frac{2x}{d}\right) \]
To find \( E(x) \):
\[ E(x) = -\frac{d}{dx} \left[-V_0 \left(1 + \frac{2x}{d}\right)\right] = V_0 \cdot \frac{2}{d} \]
Thus:
\[ E(x) = \frac{2V_0}{d} \]

- **For \( x \geq d \)**:
\[ E(x) = -\frac{d}{dx} \left[-3V_0\right] = 0 \]

3. **Summarize the Electric Field**:
- For \( x < -d \): \( E(x) = 0 \)
- For \( -d \leq x \leq 0 \): \( E(x) = \frac{2V_0}{d} \left(1 + \frac{x}{d}\right) \) (linear function)
- For \( 0 < x < d \): \( E(x) = \frac{2V_0}{d} \) (constant)
- For \( x \geq d \): \( E(x) = 0 \)

4. **Graph the Electric Field**:
- The graph will show:
- A horizontal line at \( E = 0 \) for \( x < -d \).
- A linear increase from \( 0 \) to \( \frac{2V_0}{d} \) as \( x \) goes from \( -d \) to \( 0 \).
- A constant value of \( \frac{2V_0}{d} \) from \( 0 \) to \( d \).
- A drop back to \( 0 \) for \( x > d \).