The electric field in a region is radially outwards with magnitude `E=alphar//epsilon_(0)`. IN a sphere of radius `R` centered at the origin, calculate the value of charge in coulombs `alpha=5/(pi)` if `V//m^(2)` and `R=(3/10)^(1//3)m`.

Explanation

To solve the problem, we need to calculate the charge enclosed within a sphere of radius \( R \) given the electric field \( E = \frac{\alpha r}{\epsilon_0} \). Here are the steps to find the charge:

### Step 1: Understand the Electric Field
The electric field is given by:
\[ E = \frac{\alpha r}{\epsilon_0} \]
where \( \alpha = \frac{5}{\pi} \) and \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Determine the Area of the Sphere
The surface area \( A \) of a sphere with radius \( R \) is given by:
\[ A = 4\pi R^2 \]

### Step 3: Calculate the Electric Flux
The electric flux \( \Phi_E \) through the surface of the sphere is given by:
\[ \Phi_E = \int E \cdot dA = E \cdot A \]
Since \( E \) is constant over the surface of the sphere, we can write:
\[ \Phi_E = E \cdot A = E \cdot 4\pi R^2 \]

### Step 4: Substitute the Electric Field into the Flux Equation
Substituting \( E \) into the flux equation:
\[ \Phi_E = \left(\frac{\alpha R}{\epsilon_0}\right) \cdot 4\pi R^2 \]
This simplifies to:
\[ \Phi_E = \frac{4\pi \alpha R^3}{\epsilon_0} \]

### Step 5: Relate Electric Flux to Charge Enclosed
According to Gauss's law, the electric flux is also equal to the charge enclosed \( Q \) divided by \( \epsilon_0 \):
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
Setting the two expressions for \( \Phi_E \) equal gives:
\[ \frac{4\pi \alpha R^3}{\epsilon_0} = \frac{Q}{\epsilon_0} \]

### Step 6: Solve for Charge \( Q \)
Multiplying both sides by \( \epsilon_0 \) and rearranging gives:
\[ Q = 4\pi \alpha R^3 \]

### Step 7: Substitute Values for \( \alpha \) and \( R \)
Substituting \( \alpha = \frac{5}{\pi} \) and \( R = \left(\frac{3}{10}\right)^{1/3} \):
\[ Q = 4\pi \left(\frac{5}{\pi}\right) \left(\left(\frac{3}{10}\right)^{1/3}\right)^3 \]
This simplifies to:
\[ Q = 4\pi \cdot \frac{5}{\pi} \cdot \frac{3}{10} = 4 \cdot 5 \cdot \frac{3}{10} = 6 \]

### Final Answer
Thus, the charge enclosed in coulombs is:
\[ Q = 6 \, \text{C} \]

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