This **article aims to find the electric flux in a cube having six sides**. This article uses the concept of electric flux. For a **closed gaussian surface** electric flux is given by the formula

\[\Phi_{e} = \dfrac{Q}{xi_{o}}\]

**Expert Answer**

Consider a** cube having side length** $ L $ in which a** size** $ q $ charge is placed at the center. Consider a closed **Gaussian surface**, which is a cube whose** electric flux** is $\Phi $, which is given by:

\[\Phi=\dfrac{ q } {\xi_{o}}\]

The number of lines of force arising from the charge will be divided into six walls. So the electric flux is given by:

\[\Phi =\dfrac{q}{6\xi_{o}}\]

**Part (A)**

The **electric flux** of each of the** six faces of the cube** is $\Phi = \dfrac{ q } { 6 \xi _{ o } } $.

**Electric flux** is **number of field lines passing per unit area**. The **flux through any face of the cube is equal to the total flux of the cube divided by six**.

Consider the **sides of the cube** $ L_{1}$.

Since the **electric flux depends** only on the **enclosed charge** $ q $, the flux through each surface would be the same as the previous part, even if the **cube’s dimension changes**. That is, the **electric flux** of each of the **six walls** of the cube, the length $ L_{ 1 } $ of which

\[\Phi _{1}=\dfrac{q}{6\xi_{o}}\]

**Part (B)**

The** electric flux of each of the six faces of the cube** is $\Phi _{ 1 }=\dfrac{q}{6\xi _{o}}$.

Since the **flux depends on the charge inside the closed surface,** the flux through each surface would be same as in the **previous section**, even if the **dimension changes.**

**Numerical Result**

(a)** Electric flux** $\Phi $ across each of the **six faces of the cube** is equal to $ \dfrac{ q } { 6 \xi _{ o } }$.

(b)** Flux** $ \Phi _{1} $ over the** face of the cube** if its sides were $ L_{1} $ long is equal to $\dfrac{ q } { 6 \xi _{ o } }$.

**Example**

A point charge of magnitude $Q$ is at the cube’s center with sides of length $x$. What is the electric flux $\Phi $ across each of the cube’s six faces? What would be the flux $ \Phi $ over the face of the cube if its sides were long $ x_{1}$?

**Solution**

Consider a closed **Gaussian surface**, which is a cube whose** electric flux** is $\Phi $ which is given by

\[\Phi =\dfrac{Q}{\xi _{o}}\]

The **number of lines** of force arising from the charge will be** divided into six walls.** So the **electric flux** is given by

\[\Phi =\dfrac{Q}{6\xi _{o}}\]

**Part (A)**

The **electric flux** of each of the** six faces of the cube** is $\Phi = \dfrac{Q}{6\xi _{ o }}$.

Consider the **sides of the cube** $ x_{1}$. That is, the **electric flux** of each of the **six walls** of the cube, the length $L_{1}$ of which

\[\Phi _{1}=\dfrac{Q}{6\xi _{o}}\]

**Part (B)**

The** electric flux of each of the six faces of the cube** is $\Phi _{1}=\dfrac{Q}{ 6 \xi _{o}}$.