The electric potential in a region is given as `V = -4 ar^2 + 3b`, where r is distance from the origin, a and b are constants. If the volume charge density in the region is given by `rho = n a epsilon_(0)`, then what is the value of n?

Answer

To solve the problem, we need to find the value of \( n \) given the electric potential \( V \) and the volume charge density \( \rho \).

### Step 1: Understand the given potential function
The electric potential is given by:
\[ V = -4ar^2 + 3b \]
where \( r \) is the distance from the origin, and \( a \) and \( b \) are constants.

### Step 2: Calculate the electric field \( E \)
The electric field \( E \) is related to the electric potential \( V \) by the equation:
\[ E = -\frac{dV}{dr} \]
We will differentiate \( V \) with respect to \( r \).

### Step 3: Differentiate the potential
Differentiating \( V \):
\[ \frac{dV}{dr} = \frac{d}{dr}(-4ar^2 + 3b) = -4a \cdot 2r + 0 = -8ar \]
Thus, the electric field \( E \) becomes:
\[ E = -\left(-8ar\right) = 8ar \]

### Step 4: Relate electric field to charge density
The relationship between the electric field and the volume charge density \( \rho \) is given by:
\[ \frac{dE}{dr} = \frac{\rho}{\epsilon_0} \]
Now, we need to differentiate \( E \) with respect to \( r \).

### Step 5: Differentiate the electric field
Differentiating \( E \):
\[ \frac{dE}{dr} = \frac{d}{dr}(8ar) = 8a \]

### Step 6: Set up the equation for charge density
From the relationship established earlier:
\[ 8a = \frac{\rho}{\epsilon_0} \]
Substituting the given volume charge density \( \rho = n a \epsilon_0 \):
\[ 8a = \frac{n a \epsilon_0}{\epsilon_0} \]
This simplifies to:
\[ 8a = n a \]

### Step 7: Solve for \( n \)
Assuming \( a \neq 0 \), we can divide both sides by \( a \):
\[ n = 8 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{8} \]