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. (a) `-6aepsilon_0r`
B. (b) `-24piaepsilon_0`
C. (c) `-6aepsilon_0`
D. (d) `-24piaepsilon_0r`

Explanation

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: Understand the relationship between electric potential and electric field
The electric field \( E \) can be derived from the electric potential \( \phi \) using the formula:
\[ E = -\frac{d\phi}{dr} \]

### Step 2: Differentiate the potential
Given \( \phi = ar^2 + b \), we differentiate \( \phi \) with respect to \( r \):
\[ \frac{d\phi}{dr} = \frac{d}{dr}(ar^2 + b) = 2ar \]
Thus, the electric field \( E \) becomes:
\[ E = -\frac{d\phi}{dr} = -2ar \]

### Step 3: Relate the electric field to charge density
According to Gauss's law in electrostatics, the electric field \( E \) inside a spherical charge distribution is related to the charge density \( \rho \) by:
\[ E = \frac{1}{4\pi \epsilon_0} \frac{Q_{enc}}{r^2} \]
where \( Q_{enc} \) is the charge enclosed within a radius \( r \).

### Step 4: Express the enclosed charge in terms of charge density
The charge \( Q_{enc} \) can be expressed in terms of the charge density \( \rho \) as:
\[ Q_{enc} = \rho \cdot \text{Volume} = \rho \cdot \frac{4}{3} \pi r^3 \]

### Step 5: Substitute \( Q_{enc} \) into the electric field equation
Substituting \( Q_{enc} \) into the electric field equation gives:
\[ E = \frac{1}{4\pi \epsilon_0} \frac{\rho \cdot \frac{4}{3} \pi r^3}{r^2} = \frac{\rho}{3\epsilon_0} r \]

### Step 6: Set the two expressions for \( E \) equal
Now we have two expressions for \( E \):
1. From the potential: \( E = -2ar \)
2. From Gauss's law: \( E = \frac{\rho}{3\epsilon_0} r \)

Setting these equal to each other:
\[ -2ar = \frac{\rho}{3\epsilon_0} r \]

### Step 7: Solve for charge density \( \rho \)
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ -2a = \frac{\rho}{3\epsilon_0} \]
Multiplying both sides by \( 3\epsilon_0 \) gives:
\[ \rho = -6a\epsilon_0 \]

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

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