7 electric dipoles are placed at the center of a cube. Each charge forming the dipole is ofQ units. The net flux coming out of the cube surface is __

1. Break Down the Problem

We need to find the net electric flux ($\Phi$) coming out of the surface of a cube that contains multiple electric dipoles. We will consider the following aspects:

• The nature of electric dipoles.
• The applicability of Gauss's law.

2. Relevant Concepts

According to Gauss's law, the net electric flux through a closed surface is given by:

$\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$

where $Q_{\text{enc}}$ is the total charge enclosed within the surface, and $\varepsilon_0$ is the permittivity of free space.

3. Analysis and Detail

1. Understanding Dipoles: An electric dipole consists of two equal and opposite charges. The net charge of a single dipole is zero ($Q + (-Q) = 0$).
2. Multiple Dipoles: Given that there are 7 electric dipoles, the total charge enclosed by the cube is:

$Q_{\text{enc}} = 7 \times (Q + (-Q)) = 7 \times 0 = 0$

4. Verify and Summarize

• The enclosed charge $Q_{\text{enc}}$ is zero because the charges of the dipoles cancel out.
• As a result, applying Gauss's law, we find that the net electric flux ($\Phi$) through the surface of the cube is zero:

$\Phi = \frac{0}{\varepsilon_0} = 0$

Final Answer

The net flux coming out of the cube surface is 0.