Consider an atom with atomic number Z as consisting of a positive point charge at the centre and surrounded by a distribution of negative electricity uniformly distributed within a sphere of radius R. The electric field at a point inside the atom at a distance r from the centre is

Options

A. `(Ze)/(4pi epsilon_(0))[(1)/(r^(2))-( r )/(R^(3))]`
B. `(Ze)/(4pi epsilon_(0))[(1)/(r^(2))+(1)/(R^(3))]`
C. `(2Ze)/(4pi epsilon_(0)r^(2))`
D. Zero

Answer

To find the electric field at a point inside an atom at a distance \( r \) from the center, we can use Gauss's law. Here’s a step-by-step solution:

### Step 1: Understand the System
We have an atom with atomic number \( Z \), which means it has \( Z \) protons (positive charge) at the center and \( Z \) electrons (negative charge) uniformly distributed within a sphere of radius \( R \).

### Step 2: Define the Gaussian Surface
To find the electric field at a distance \( r \) from the center, we consider a spherical Gaussian surface of radius \( r \) (where \( r < R \)).

### Step 3: Apply Gauss's Law
Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\( \epsilon_0 \)):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux.

### Step 4: Calculate the Enclosed Charge
1. **Positive Charge**: The charge at the center is \( +Ze \).
2. **Negative Charge**: The negative charge is distributed uniformly throughout the volume of the atom. The total negative charge is \( -Ze \).

The volume charge density \( \rho \) of the negative charge is:
\[ \rho = \frac{-Ze}{\frac{4}{3} \pi R^3} \]
The charge enclosed within the Gaussian surface of radius \( r \) is:
\[ Q_{\text{enc}} = \rho \cdot \text{Volume of sphere of radius } r = \left(\frac{-Ze}{\frac{4}{3} \pi R^3}\right) \cdot \left(\frac{4}{3} \pi r^3\right) = -Ze \cdot \frac{r^3}{R^3} \]
Thus, the total enclosed charge \( Q_{\text{enc}} \) becomes:
\[ Q_{\text{enc}} = Ze - Ze \cdot \frac{r^3}{R^3} = Ze \left(1 - \frac{r^3}{R^3}\right) \]

### Step 5: Calculate the Electric Flux
The electric field \( E \) is uniform over the Gaussian surface, so:
\[ \Phi_E = E \cdot 4\pi r^2 \]
Substituting into Gauss's law:
\[ E \cdot 4\pi r^2 = \frac{Ze \left(1 - \frac{r^3}{R^3}\right)}{\epsilon_0} \]

### Step 6: Solve for Electric Field \( E \)
Rearranging the equation:
\[ E = \frac{Ze \left(1 - \frac{r^3}{R^3}\right)}{4\pi \epsilon_0 r^2} \]

### Final Result
The electric field at a point inside the atom at a distance \( r \) from the center is:
\[ E = \frac{Ze}{4\pi \epsilon_0} \left(\frac{1}{r^2} - \frac{r}{R^3}\right) \]