The electrostatic potential inside a charged spherical ball is given by `phi=ar^2+b` where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is:

Options

A. `-24 pi a epsi_(0) r`
B. `-6pi a epsi_(0) r`
C. `-24 pi a epsi_(0)`
D. `-6 a epsi_(0)`

Answer

To find the charge density inside a charged spherical ball where the electrostatic potential is given by \( \phi = ar^2 + b \), we can follow these steps:

### Step 1: Find the Electric Field
The electric field \( E \) is related to the electrostatic potential \( \phi \) by the relation:
\[ E = -\frac{d\phi}{dr} \]
Given \( \phi = ar^2 + b \), we differentiate it with respect to \( r \):
\[ E = -\frac{d}{dr}(ar^2 + b) = -\frac{d}{dr}(ar^2) - \frac{d}{dr}(b) = -2ar \]
Thus, the electric field inside the charged spherical ball is:
\[ E = -2ar \]

### Step 2: Relate Electric Field to Charge Density
Using Gauss's law, we know that the electric field \( E \) is also related to the charge density \( \rho \) by the equation:
\[ E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \]
where \( Q \) is the charge enclosed within a radius \( r \). The charge density \( \rho \) can be expressed as:
\[ \rho = \frac{dQ}{dV} \]
For a spherical volume, the volume element \( dV \) is given by \( dV = 4\pi r^2 dr \).

### Step 3: Calculate Charge Enclosed
From the electric field expression, we can find the total charge \( Q \) enclosed within radius \( r \):
\[ E = -2ar \implies -2ar = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \]
Rearranging gives:
\[ Q = -8\pi\epsilon_0 ar^3 \]

### Step 4: Differentiate Charge to Find Charge Density
Now, we differentiate \( Q \) with respect to \( r \) to find the charge density \( \rho \):
\[ \frac{dQ}{dr} = \frac{d}{dr}(-8\pi\epsilon_0 ar^3) = -24\pi\epsilon_0 ar^2 \]
Thus, the charge density \( \rho \) is:
\[ \rho = \frac{dQ}{dV} = \frac{dQ}{dr} \cdot \frac{1}{dV/dr} = \frac{-24\pi\epsilon_0 ar^2}{4\pi r^2} = -6\epsilon_0 a \]

### Conclusion
The charge density inside the ball is:
\[ \rho = -6\epsilon_0 a \]