The electric potential (V) in a certain region of space depends only on x-coordinate of point as `V=-alpha x^(3)+beta` (`alpha and beta` constants). Find the volume charge density `(rho)` of this region of space

Answer

To find the volume charge density \( \rho \) in the given region of space where the electric potential \( V \) depends only on the x-coordinate, we can follow these steps:

### Step 1: Write down the expression for electric potential
The electric potential \( V \) is given as:
\[ V = -\alpha x^3 + \beta \]
where \( \alpha \) and \( \beta \) are constants.

### Step 2: Find the electric field \( E \)
The electric field \( E \) is related to the electric potential \( V \) by the equation:
\[ E = -\frac{dV}{dx} \]
We need to differentiate \( V \) with respect to \( x \):
\[ E = -\frac{d}{dx}(-\alpha x^3 + \beta) = -(-3\alpha x^2) = 3\alpha x^2 \]

### Step 3: Use Gauss's law to relate electric field and charge density
According to Gauss's law, the divergence of the electric field is related to the charge density \( \rho \) by:
\[ \nabla \cdot E = \frac{\rho}{\epsilon_0} \]
In one dimension (along the x-axis), this simplifies to:
\[ \frac{dE}{dx} = \frac{\rho}{\epsilon_0} \]

### Step 4: Differentiate the electric field with respect to \( x \)
Now we differentiate \( E \) with respect to \( x \):
\[ \frac{dE}{dx} = \frac{d}{dx}(3\alpha x^2) = 6\alpha x \]

### Step 5: Substitute into Gauss's law
Now we substitute \( \frac{dE}{dx} \) into the equation from Gauss's law:
\[ 6\alpha x = \frac{\rho}{\epsilon_0} \]

### Step 6: Solve for volume charge density \( \rho \)
Rearranging the equation gives us the expression for the volume charge density \( \rho \):
\[ \rho = 6\alpha x \epsilon_0 \]

### Final Answer
Thus, the volume charge density \( \rho \) in this region of space is:
\[ \rho = 6\alpha x \epsilon_0 \]

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