banner

A slab of insulating material of uniform thickness d, lying between -d/2 to d/2 along the x-axis, extends infinitely in the y and z directions. The slab has a uniform charge density p. The electric field is zero in the middle of the slab, at x=0. What is ein(x), the magnitude of the electric field inside the slab as a function of x?

  • 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 $1485 Vm$.

5/5 - (17 votes)