Question

What is electric flux? Write its S.I. units. Using Gauss’s theorem, deduce an expression for the electric field at a point due to a uniformly charged infinite plane sheet.

Answer 1

In electromagnetism, electric flux is the measure of flow of the electricfield through a given area. Electric flux is proportional to the number of electric field lines going through a normally perpendicular surface. If the electric field is uniform, the electric flux passing through a surface of vector area S is.

Answer 2

(i) Electric flux over an area in an electric field represents the total number of electric lines of force crossing the area in a direction normal to the plane of the area. The SI unit of electric flux is Nm²/C

(ii) Gauss' law states that the total electric flux through a closed surface $\displaystyle\sf = \dfrac{1}{\varepsilon_0}$ times, the magnitude of the change enclosed by it is

$\displaystyle\sf \phi = \dfrac{q}{\varepsilon_0}$

Here ε_0 is the absolute permittivity of the free space and q is the total charge enclosed

Also, $\displaystyle\sf \phi = \oint_s \vec{\bf E}\sf \cdot \vec{d\bf S} \sf = \dfrac{q}{\varepsilon_0}$

where E is the electric field at the area element dS.

Now, the electric field $\displaystyle\bf \vec{E} = \sf Cx \hat{i}$ is in X-direction only. So, faces with surface normal vector perpendicular to this field would give zero electric flux, i.e. $\displaystyle\sf \phi = E\:dS\:\cos 90^{\circ}$ through it.

[refer to attachment (1)].

So, flux would be across only two surfaces.

Magnitude of E at left face,

$\displaystyle\sf E_L = Cx = Ca$ [x = a ay left side]

Magnitude of E at right face,

$\displaystyle\sf E_R = Cx = C2a = 2aC$ [x = 2a at right face]

Thus corresponding fluxes are

$\displaystyle\sf \phi_L = \bf \vec{E_{\sf L}}\cdot\sf\vec{ d \bf S}$

$\displaystyle\sf = E_L\:dS\:\cos\theta$

$\displaystyle\sf = - aC \times a^2$ [θ = 180°]

$\displaystyle\sf$

$\displaystyle\sf \phi_R = \bf \vec{E_{\sf R}}\sf\cdot\vec{ d\bf S}$

$\displaystyle\sf = 2aC\:ds\:\cos\theta$

$\displaystyle\sf = 2aCa^2 = 2a^3 C$ [θ = 0°]

$\displaystyle\sf$

• now, net flux through cube

$\displaystyle\sf = \phi_L + \phi_R$

$\displaystyle\sf = -a^3C + 2a^3C$

$\displaystyle\sf = a^3C\:Nm^2\:C^{-1}$

$\displaystyle\sf$

• net charge inside cube

$\displaystyle\sf \phi = \dfrac{q}{\varepsilon_0}$

$\displaystyle\sf q = \phi \varepsilon_0$

$\displaystyle\sf q = a^3C\varepsilon_0$ coulomb

$\displaystyle\sf$

Now, let us consider a large plane sheet or charge density sigma

[refer to attachment (2)].

Let the electric field is to be obtained at a point P at a distance r from it.

Now, applying Gauss' law over the closed gaussian surface

$\displaystyle\sf \oint_S \bf \vec{E} \cdot \sf \vec{d\bf S} = \sf \dfrac{q}{\varepsilon_0}$

$\displaystyle\sf \int_{CSA} \bf\vec{ E} \cdot \sf \vec{d\bf S} \sf + \int_{CSA} \bf \vec{E} \cdot \sf \vec{d\bf S} \sf= \dfrac{q}{\varepsilon_0}$

$\displaystyle\sf \int_{CSA} E\:dS\:\cos 90^{\circ} + \int E\:dS\:\cos 90^{\circ} = \dfrac{q}{\varepsilon_0}$

$\displaystyle\sf \int_{CSA} dS = \dfrac{q}{\varepsilon_0}$

$\displaystyle\sf \implies E \times 2A = \dfrac{q}{\varepsilon_0} \;\;\;\;\dots (i)$

$\displaystyle\sf E = \dfrac{q}{2A\varepsilon_0} = \dfrac{\sigma}{2\varepsilon_0}$ [ from eqn. (1)].

$\displaystyle\boxed{\sf \therefore\text{electric field intensity} = \dfrac{\sigma}{2\varepsilon_0}}$

The direction of E is normal to the plane sheet and directed away from sheet when charge on plane is positive and vice-versa.

• CS comprises of two caps and Curved surface area.

Now, The field directed,

• Normally away from the sheet when sheet is positively charged.
• Normally inward towards the sheet when plane sheet is negatively charged.