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 \( V = -\frac{x^3}{6\epsilon_0} + 2 \) volts, we can follow these steps:

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

### Step 2: Differentiate the Potential
The potential \( V \) is given as:
\[ V = -\frac{x^3}{6\epsilon_0} + 2 \]
Now, we differentiate \( V \):
\[ \frac{dV}{dx} = -\frac{d}{dx}\left(\frac{x^3}{6\epsilon_0}\right) + \frac{d}{dx}(2) \]
The derivative of a constant is zero, and the derivative of \( -\frac{x^3}{6\epsilon_0} \) is:
\[ -\frac{3x^2}{6\epsilon_0} = -\frac{x^2}{2\epsilon_0} \]
Thus,
\[ \frac{dV}{dx} = -\frac{x^2}{2\epsilon_0} \]
Now substituting this into the equation for \( E \):
\[ E = -\left(-\frac{x^2}{2\epsilon_0}\right) = \frac{x^2}{2\epsilon_0} \]

### Step 3: Relate Electric Field to Charge Density
Using Gauss's law, the charge density \( \rho \) can be found from the electric field:
\[ \rho = \epsilon_0 \frac{dE}{dx} \]
Now, we need to differentiate \( E \) with respect to \( x \):
\[ E = \frac{x^2}{2\epsilon_0} \]
Differentiating \( E \):
\[ \frac{dE}{dx} = \frac{d}{dx}\left(\frac{x^2}{2\epsilon_0}\right) = \frac{2x}{2\epsilon_0} = \frac{x}{\epsilon_0} \]

### Step 4: Calculate Charge Density
Now substituting \( \frac{dE}{dx} \) into the equation for \( \rho \):
\[ \rho = \epsilon_0 \cdot \frac{x}{\epsilon_0} = x \]

### Step 5: Evaluate Charge Density 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:
\[ \rho = 2 \, \text{C/m}^3 \]