**Find the equation of $E_{out}$, the magnitude of the electric field outside the slab.****Find the equation of $E_{in}$, the magnitude of the electric field inside the slab.**

This question aims to find the **electric field inside** and **outside** of an **insulating slab** lying on the **cartesian plane.**

This question is based on the concept of **Gauss’s Law, electric field,** and **electric flux. Electric flux** can be defined as the **number** of **lines** of **electric force** passing through an **area** of a **surface****.**

## Expert Answer

**a)** Calculate the **magnitude** of the **electric field outside** the **slab** by using the **electric flux** formula given by **Gauss’s Law** as:

\[ Electric\ Flux\ \Phi\ =\ A \times E_ {out} \]

**Electric flux** is also equal to the **total charge** over **dielectric permittivity** of **vacuum** by **superposition principle,** which is given as:

\[ Electric\ Flux\ \Phi\ =\ \dfrac {Q} { \varepsilon_0} \]

As the total **electric flux outside** the whole slab will be the same, we can write these equations as:

\[ E_{out}\ A = \dfrac {Q} {|varepsilon_0} \]

Solving for the **electric field outside** the **slab,** we get:

\[ E_{out}\ A = \dfrac { A\ \rho\ d} {2 \varepsilon_0} \]

\[ E_{out} = \dfrac {d \rho} {2 \varepsilon_0} \]

**b)** Using the formula for **electric flux** given by the **Gauss’s Law** and **superposition principle** as:

\[ E_{in}\ A = \dfrac {Q} {\varepsilon_0} \]

Substituting the value of $Q$, we can calculate the expression for the **magnitude** of the **electric field inside** the **slab** as:

\[ E_{in}\ A = \dfrac{A\ \rho\ X} {\varepsilon_0} \]

\[ E_{in}\ = \dfrac{ \rho\ } { \varepsilon_0} X \]

## Numerical Result

**a)** The **magnitude** of the **electric field outside** the given **slab** is calculated to be:

\[ E_{out} = \dfrac {d\ \rho} {2 \varepsilon_0} \]

**b)** The **magnitude** of the **electric field inside** the given **slab** is calculated to be:

\[ E_{in}\ = \dfrac{ \rho } { \varepsilon_0} X \]

## Example

Find the **electric flux** that passes through a **sphere** which an **electric field** of $1.5k V/m$ and makes **angle** of $45^{\circ}$ with **surface vector** of the **sphere. Area** of the **sphere** is given as $1.4 m^2$.

The given information about the question is as follows:

\[ Electric\ Field\ E\ =\ 1500 V/m \]

\[ Area\ of\ the\ Sphere\ A\ =\ 1.4 m^2 \]

\[ Angle\ \theta\ =\ 45^{\circ} \]

To calculate the **electric flux,** we can use the formula by **Gauss’s Law**:

\[ \Phi = E.A \]

\[ \Phi = E A \cos \theta \]

\[ \Phi = (1500 V/m) (1.4 m^2) \cos(45 ^{\circ}) \]

Solving the equation will give us:

\[ \Phi = 1485 V m \]

The **electric flux** of the given problem is calculated to be $1485 Vm$.