The electrostatic potential existing in the space is given as `V=-((x^(3))/(6epsilon_(0))+2)` volts. Find the charge density (in `"coulomb"//m^(3)`) at `x=2m`.

Explanation

To find the charge density at \( x = 2 \, \text{m} \) given the electrostatic potential \( V = -\frac{x^3}{6\epsilon_0} + 2 \), we will follow these steps:

### Step 1: Find the Electric Field
The electric field \( E \) is related to the electrostatic potential \( V \) by the formula:
\[ E = -\frac{dV}{dx} \]
We need to differentiate \( V \) with respect to \( x \).

### Step 2: Differentiate the Potential
Given:
\[ V = -\frac{x^3}{6\epsilon_0} + 2 \]
Differentiating \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = -\frac{1}{6\epsilon_0} \cdot 3x^2 = -\frac{x^2}{2\epsilon_0} \]
Thus, the electric field \( E \) becomes:
\[ E = -\left(-\frac{x^2}{2\epsilon_0}\right) = \frac{x^2}{2\epsilon_0} \]

### Step 3: Relate Electric Field to Charge Density
The charge density \( \rho \) can be found using Gauss's law, which relates the divergence of the electric field to the charge density:
\[ \rho = \epsilon_0 \nabla \cdot E \]
In one dimension, this simplifies to:
\[ \rho = \epsilon_0 \frac{dE}{dx} \]

### Step 4: Differentiate the Electric Field
Now we differentiate \( E \) with respect to \( x \):
\[ E = \frac{x^2}{2\epsilon_0} \]
Differentiating \( E \):
\[ \frac{dE}{dx} = \frac{1}{2\epsilon_0} \cdot 2x = \frac{x}{\epsilon_0} \]

### Step 5: Find the Charge Density
Substituting \( \frac{dE}{dx} \) back into the equation for charge density:
\[ \rho = \epsilon_0 \cdot \frac{x}{\epsilon_0} = x \]

### Step 6: Evaluate at \( x = 2 \, \text{m} \)
Now we can find the charge density at \( x = 2 \, \text{m} \):
\[ \rho(2) = 2 \, \text{C/m}^3 \]

### Final Answer
The charge density at \( x = 2 \, \text{m} \) is:
\[ \rho = 2 \, \text{C/m}^3 \]

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