• What is the net electric flux through a closed surface surrounding an electric dipole? Derive the expression for electric field intensity both inside and outside a uniformly charged spherical call. What is the total charged enclosed by a closed surface if the electric flux entering and leaving the surface are 20000 N / C^m2 and 30000 N / C^m2 respectively? Given Epsilon, not = 8.85* 10^-12C²N^-1m^-2.

To address your question about the net electric flux through a closed surface surrounding an electric dipole, as well as deriving the expression for electric field intensity around a uniformly charged spherical shell, let's break down these concepts step by step. Additionally, we will calculate the total charge enclosed by a surface given specific electric flux values.

Electric Flux and Electric Dipoles

The net electric flux (\( \Phi_E \)) through a closed surface surrounding an electric dipole is zero. This is rooted in Gauss's Law, which states that the electric flux through a closed surface is proportional to the charge enclosed by that surface:

Gauss's Law: \( \Phi_E = \frac{Q_{enc}}{\epsilon_0} \)

For an electric dipole, which consists of two equal and opposite charges separated by a distance, the total enclosed charge (\( Q_{enc} \)) is zero. Thus, the net electric flux through a surface enclosing the dipole is:

\( \Phi_E = \frac{0}{\epsilon_0} = 0 \)

Electric Field Intensity of a Uniformly Charged Spherical Shell

Next, let's derive the expression for the electric field intensity both inside and outside a uniformly charged spherical shell.

Outside the Spherical Shell

For a uniformly charged spherical shell with total charge \( Q \) and radius \( R \), the electric field (\( E \)) at a distance \( r \) from the center (where \( r > R \)) can be derived from Gauss's Law. We choose a Gaussian surface that is a sphere of radius \( r \) (greater than \( R \)). The symmetry implies that the electric field is constant on this surface:

According to Gauss's Law:

\( \Phi_E = E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \)

Solving for \( E \), we get:

Electric Field Outside: \( E = \frac{Q}{4\pi \epsilon_0 r^2} \)

Inside the Spherical Shell

For points inside the uniformly charged spherical shell (where \( r < R \)), we consider a Gaussian surface of radius \( r \). Since there is no charge enclosed within this Gaussian surface, we apply Gauss's Law:

\( \Phi_E = E \cdot 4\pi r^2 = 0 \)

Thus, the electric field inside the shell is:

Electric Field Inside: \( E = 0 \)

Calculating Total Charge Enclosed from Electric Flux

Now, let's determine the total charge enclosed by a closed surface when the electric flux entering and leaving the surface is given. The electric flux entering the surface is \( 20000 \, \text{N/C m}^2 \) and the flux leaving is \( 30000 \, \text{N/C m}^2 \). The net electric flux (\( \Phi_E \)) through the surface is:

\( \Phi_E = \text{Flux leaving} - \text{Flux entering} \)

\( \Phi_E = 30000 \, \text{N/C m}^2 - 20000 \, \text{N/C m}^2 = 10000 \, \text{N/C m}^2 \)

Using Gauss's Law to find the total charge enclosed (\( Q_{enc} \)):

\( Q_{enc} = \Phi_E \cdot \epsilon_0 \)

Substituting the value of \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):

\( Q_{enc} = 10000 \, \text{N/C m}^2 \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Calculating this gives:

\( Q_{enc} = 8.85 \times 10^{-8} \, \text{C} \)

This represents the total charge enclosed by the closed surface. Each of these concepts—electric flux, electric field intensity, and charge—are foundational in understanding electrostatics and play a significant role in various applications in physics and engineering.