The field potentail in a certain region of space depends only on the `x` coordinate as `varphi = -ax^(2) + b`, where `a` and `b` are constants. Find the distribution of the space charge `rho (x)`.

Answer

To find the distribution of space charge \(\rho(x)\) given the electric potential \(\varphi = -ax^2 + b\), we will follow these steps:

### Step 1: Write down the expression for electric potential
The electric potential is given as:
\[ \varphi(x) = -ax^2 + b \]

### Step 2: Calculate the electric field \(E(x)\)
The electric field \(E\) is related to the electric potential \(\varphi\) by the equation:
\[ E = -\frac{d\varphi}{dx} \]
We will differentiate \(\varphi\) with respect to \(x\):
\[ E(x) = -\frac{d}{dx}(-ax^2 + b) = -(-2ax) = 2ax \]
Thus, the electric field is:
\[ E(x) = 2ax \]

### Step 3: Apply Gauss's law in differential form
According to Gauss's law in differential form, the divergence of the electric field is related to the charge density \(\rho(x)\):
\[ \frac{dE}{dx} = \frac{\rho(x)}{\epsilon_0} \]
We will differentiate \(E(x)\) with respect to \(x\):
\[ \frac{dE}{dx} = \frac{d}{dx}(2ax) = 2a \]

### Step 4: Relate the electric field to charge density
Now, substituting \(dE/dx\) into Gauss's law:
\[ 2a = \frac{\rho(x)}{\epsilon_0} \]
Rearranging this gives us the expression for the charge density:
\[ \rho(x) = 2a \epsilon_0 \]

### Final Answer
The distribution of the space charge \(\rho(x)\) is:
\[ \rho(x) = 2a \epsilon_0 \]