The field potential inside a charged ball of radius R and centre at O depends only on distance from its centre as `V( r) = alphar^(2) + beta` when `alpha , beta` are `+ ve` constant . Now choose correct options

Options

A. Electric field inside the ball `E_(r ) = - 2 alpha r`
B. Electric flux passing through an imaginary sphere of radius r centre at O will be `-2 pi alphar^(2)`
C. Volume charge density ` rho (r )` inside ball is `- epsilon_(0)alpha`
D. Electric energy of charges ball will be `( 48 )/ 5 pi epsilon_(0) alpha^(2) R^(5)`

Answer

To solve the problem, we need to analyze the electric potential \( V(r) \) given by the equation:

\[ V(r) = \alpha r^2 + \beta \]

where \( \alpha \) and \( \beta \) are positive constants. We will find the electric field \( E \) inside the charged ball using the relationship between electric potential and electric field.

### Step 1: Understand the relationship between electric potential and electric field

The electric field \( E \) is related to the electric potential \( V \) by the formula:

\[ E = -\frac{dV}{dr} \]

This means that to find the electric field, we need to take the derivative of the potential with respect to the distance \( r \).

### Step 2: Differentiate the potential function

Now we will differentiate the given potential function \( V(r) \):

\[ V(r) = \alpha r^2 + \beta \]

Taking the derivative with respect to \( r \):

\[ \frac{dV}{dr} = \frac{d}{dr}(\alpha r^2 + \beta) = 2\alpha r + 0 = 2\alpha r \]

### Step 3: Calculate the electric field

Now, substituting this result into the formula for the electric field:

\[ E = -\frac{dV}{dr} = -2\alpha r \]

### Step 4: Analyze the result

From our calculation, we find that the electric field inside the charged ball is:

\[ E = -2\alpha r \]

Since \( \alpha \) is a positive constant, the electric field \( E \) is directed towards the center of the ball (indicated by the negative sign) and its magnitude increases linearly with \( r \).

### Conclusion

The electric field inside the charged ball is given by:

\[ E = -2\alpha r \]