Electric Flux and Area Vector for JEE

An electric field is a region around a charged object or a charged particle within which the effect of electric force is present. This is analogous to the gravitational field or magnetic field. The electric field has electric field lines. These are the vector field lines that describe the acceleration vector at a particular point.

According to the electric flux notes PDF, electric flux  is the measure of the number of electric field lines passing through a given area of cross-section. Electric flux is the cross-product of the electric field vector and the area vector. The area vector describes the vector which is perpendicular to the plane of the area of the cross-section. Its magnitude is the value of the area of the cross-section and its direction is perpendicular to the cross-sectional plane.

Area Vector and Its Direction

Consider a finite flat plate or a finite plane, in 3-D space, oriented in any manner. Let the plane be a square and have side ‘s’. The area will be A = s². Now, the formula of the area is a cross-product between two vectors that describe the adjacent sides of the figure. This is the reason why the area is a vector quantity.

Finite Flat Plate of Side ‘s’

Since the direction of the area vector of the cross-product vector is mutually perpendicular to the two vectors, the area vector is perpendicular to the sides of the cross-sectional plane. In the above case the area vector will be $\overrightarrow{A}=A\hat{n}$, where $\hat{n}$ is the unit vector perpendicular to the plane.

Electric Flux

Electric flux is a measure of the number of electric field lines passing through a given area of cross-section. Mathematically, it is the dot product of the electric field vector and the area vector. When the field lines are entering the closed surface, then the flux is taken as negative. This is because the angle between the area vector and the electric field vector is 180⁰. When the field lines are emerging out of the closed surface, the flux is taken as positive because the angle between the vectors is 0⁰.

The formula of electric flux is $\phi =\vec{B}.\vec{A}$.

When the electric field lines are passing through the plane of the cross-section at an angle, the components of the electric field are taken. One component is along the area vector and the other one is perpendicular to the area vector. So, the component that is along the area vector is taken and the flux is calculated accordingly.

Electric Flux Density Definition

When the electric field lines pass through a unit area, it is called electric field density. Electric flux density is a measure of the strength of an electric field generated by a free electric charge, corresponding to the number of electric lines of force passing through a given area.

The formula of electric flux density is $D={\varepsilon }_{0} E$.

Units of Electric Flux and Area

Mathematically, electric flux is the product of electric field and area. The units of the electric field are Newton per Coulomb or N/C. The units of area are m². So, the units of electric flux come out to be Nm²/C. We know that $\dfrac{Nm}{C}=\dfrac{J}{C}=V$(Volt). So, the units get reduced to Vm (Voltmeter). In the basic form, units of electric flux can be:

$\dfrac{N{{m}^{2}}}{C}=\dfrac{kg{{m}^{3}}}{{{s}^{2}}C}$

$\dfrac{N{{m}^{2}}}{C}=\dfrac{kg{{m}^{3}}s}{{{s}^{3}}C}$

$\dfrac{N{{m}^{2}}}{C}=\dfrac{kg{{m}^{3}}}{{{s}^{3}}A}$

$\dfrac{N{{m}^{2}}}{C}=kg{{m}^{3}}{{s}^{-3}}{{A}^{-1}}$

The dimensional formula is as follows:

Dimensions of the electric field are [M¹L¹I^-1T^-3].

Dimensions of area are [L²].

Thus, the dimensions of electric flux are [M¹L¹I^-1T^-3].[L²] =[ M¹L³I^-1T^-3] 

Relation between Electric Flux Density and Electric Field Intensity

The electric flux density (D) is directly proportional to the electric field intensity (E) or electric field. The relation between them is $\overrightarrow{D}={{\varepsilon }_{0}}\overrightarrow{E}$.

Consider a sphere of charge Q. The electric field at a radius R is given by $\overrightarrow{E}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{R}^{2}}}\hat{R}$

Now, integrate both the sides over the surface S of radius R.

$\oint_{S}{E\left( r \right).ds}=\oint_{S}{\left[ \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{R}^{2}}}\hat{R} \right]}.ds$

$\oint_{S}{E\left( r \right).ds}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{R}^{2}}}\oint_{S}{{\hat{R}}}.ds$

Now, $ds=\hat{R}ds$ and $\hat{R}.\hat{R}=1$. The right-hand side will become

$\oint_{S}{E\left( r \right).ds}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{R}^{2}}}\oint_{S}{ds}$

$\oint_{S}{E\left( r \right).ds}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{R}^{2}}}\times 4\pi {{R}^{2}}$

$\oint_{S}{E\left( r \right).ds}=\dfrac{Q}{{{\varepsilon }_{0}}}$

Now, the electric field at the radius R will be the same all over the sphere.

$E(r)\oint_{S}{ds}=\dfrac{Q}{{{\varepsilon }_{0}}}$

$E(r)\times 4\pi {{R}^{2}}=\dfrac{Q}{{{\varepsilon }_{0}}}$

$E=\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{Q}{4\pi {{R}^{2}}}$

The electric flux density is the enclosed charge divided by the area of the sphere.

$E=\dfrac{1}{{{\varepsilon }_{0}}}D$

$\overrightarrow{D}={{\varepsilon }_{0}}\overrightarrow{E}$

The electric flux density vector has the same direction as the electric field intensity. The units of electric flux density are C/m² and the dimensional formula is M¹L³I^-1T^-3.

Conclusion

Electric flux is a measure of the number of electric field lines passing through a given area of cross-section. Mathematically, it is the dot product of the electric field vector and the area vector. If the vectors are along the same direction, the flux is positive but when the vectors are opposite to each other, the flux is negative. The units of electric flux are Nm²/C. 

The relation between electric flux and electric field intensity is $E=\varepsilon D$. The flux is influenced by the charges present around. More the charges, the more the flux. If the same amount of charge produces the same number of field lines, the electric flux through any closed surface (that includes the charge) must be the same. Moreover, it doesn't matter where you keep the charge inside the closed surface, the number of field lines passing through the surface would be the same.