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-by-step Solution:

1. **Identify the Electric Potential**:
The electric potential is given by:
\[ V = -4a r^2 + 3b \]
where \( r \) is the distance from the origin, and \( a \) and \( b \) are constants.

2. **Calculate the Electric Field**:
The electric field \( \mathbf{E} \) can be found from the electric potential using the relation:
\[ \mathbf{E} = -\nabla V \]
In spherical coordinates, the electric field in the radial direction is given by:
\[ E_r = -\frac{dV}{dr} \]

Now, we differentiate \( V \) with respect to \( r \):
\[ \frac{dV}{dr} = \frac{d}{dr}(-4a r^2 + 3b) = -8a r \]
Therefore, the electric field is:
\[ E_r = -(-8a r) = 8a r \]

3. **Apply Gauss's Law**:
According to Gauss's law, the electric field through a closed surface is related to the charge enclosed:
\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a spherical surface of radius \( r \):
\[ \Phi_E = E_r \cdot A = (8a r) \cdot (4\pi r^2) = 32\pi a r^3 \]
Thus, we have:
\[ \frac{Q_{\text{enc}}}{\epsilon_0} = 32\pi a r^3 \]

4. **Express Charge in Terms of Volume Charge Density**:
The total charge \( Q_{\text{enc}} \) can also be expressed in terms of the volume charge density \( \rho \):
\[ Q_{\text{enc}} = \rho \cdot V = \rho \cdot \left(\frac{4}{3}\pi r^3\right) \]
Given that \( \rho = n a \epsilon_0 \), we can substitute:
\[ Q_{\text{enc}} = (n a \epsilon_0) \cdot \left(\frac{4}{3}\pi r^3\right) \]

5. **Set the Two Expressions for Charge Equal**:
Equating the two expressions for \( Q_{\text{enc}} \):
\[ n a \epsilon_0 \cdot \left(\frac{4}{3}\pi r^3\right) = 32\pi a r^3 \]

6. **Solve for \( n \)**:
Cancel \( a \) and \( r^3 \) from both sides (assuming \( a \neq 0 \) and \( r \neq 0 \)):
\[ n \epsilon_0 \cdot \frac{4}{3} = 32 \]
Rearranging gives:
\[ n = \frac{32 \cdot 3}{4 \epsilon_0} = \frac{96}{4} = 24 \]

### Final Answer:
Thus, the value of \( n \) is:
\[ \boxed{24} \]