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Step 1: Understand the Problem
We have the electric potential given by
V = -4\,a\,r^2 + 3\,b,
where r is the distance from the origin, and a and b are constants. The volume charge density is expressed as
\rho = n\,a\,\epsilon_0.
We need to find the numerical value of n .
Step 2: Relation Between Electric Field and Potential
The electric field \mathbf{E} is the negative gradient of the potential:
\mathbf{E} = -\nabla V.
Since V depends only on the radial distance r , in spherical coordinates this reduces to:
E(r) = -\frac{dV}{dr}.
Differentiate V = -4\,a\,r^2 + 3\,b with respect to r :
\frac{dV}{dr} = -8\,a\,r.
Therefore,
E(r) = -\bigl(-8\,a\,r\bigr) = 8\,a\,r.
Step 3: Apply Gauss’s Law
Gauss’s law states:
\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}.
For a spherical surface of radius r , the left side (electric flux) becomes:
\Phi_E = E(r) \times 4\pi r^2 = (8\,a\,r)\times 4\pi r^2 = 32\,\pi\,a\,r^3.
Step 4: Enclosed Charge for Uniform Density
If the volume charge density is uniform and equal to \rho , the total enclosed charge within radius r is:
Q_{\text{enc}} = \rho \times \text{Volume of sphere} = \rho \times \frac{4}{3}\,\pi r^3.
According to Gauss’s law:
32\,\pi\,a\,r^3 = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{\rho \,\bigl(\tfrac{4}{3}\,\pi r^3\bigr)}{\epsilon_0}.
Step 5: Solve for \rho
Cancel \pi r^3 on both sides:
32\,a = \frac{\rho \,\bigl(\tfrac{4}{3}\bigr)}{\epsilon_0}.
Hence,
\rho = \frac{32\,a \times 3}{4}\,\epsilon_0 = 24\,a\,\epsilon_0.
Step 6: Identify n
Since \rho is given to be n\,a\,\epsilon_0 , we match:
n\,a\,\epsilon_0 = 24\,a\,\epsilon_0 \quad\Longrightarrow\quad n = 24.
Final Answer
n = 24.
