Gauss’s Law : Introduction | Electric Flux | Applications | Examples

Electric lines of Force are used to measure Electric Flux. It is a property of Electric Field which tells us the number of field lines crossing a particular area. It is rate of flow of electric field through a surface which can be open or closed. We will also discuss Gauss’s Law in detail which is an application of Electric Flux which helps to calculate Electric Field for a given charge distribution enclosed by a closed surface. We will study its application in detail here. Here you will get to know in detail about:

1. Introduction to Electric Flux
2. Important Points on Electric Flux
3. Gauss’s Law
4. Important Points on Gauss’s Law
5. Application of Gauss’s Law
6. Examples based on Gauss’s Law

Introduction to Electric Flux

Electric flux through an elementary area, ds is defined as the scalar product of area and field, i.e. ${\varphi _{\rm{E}}} = \int {\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to } $ . It represents the total lines of force passing through the given area. Here area is treated as a vector. The direction of area vector is given by direction of normal to the surface.

Important Points on Electric Flux

1. It is a real scalar physical quantity with unit volt × m and dimensions ${\varphi _E} = Eds = {F \over q}ds = {{ML{T^{ - 2}}} \over {AT}}{L^2} = M{L^3}{T^{ - 3}}{A^{ - 1}}$ [caption id="attachment_4412" align="aligncenter" width="450"] Electric Flux varying with Electric field[/caption]
2. It will be maximum when cos θ is max = 1, i.e., θ = 0°, i.e., electric field is normal to the area with (dΦ_E)max = E ds.
3. It will be minimum when cos θ is min = 0, i.e., θ = 90°, i.e. field is parallel to the area with (dΦ_E)min = 0.
4. For a closed body outward flux is taken as positive while inward flux is taken as negative.

[caption id="attachment_4413" align="aligncenter" width="522"] For a closed body outward flux is taken as positive while inward flux is taken as negative[/caption]

Gauss's Law

Gauss’s law for electricity states that the electric flux across any closed surface is proportional to the net electric charge enclosed by the surface. The law implies that isolated electric charges exist and that like charges repel one another while unlike charges attract. [source] It relates the total flux of an electric field through a closed surface to the net charge enclosed by that surface. According to it, the total flux linked with a closed surface is 1/ε₀ times the charge enclosed by the closed surface, Mathematically$\oint\limits_s {\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to } = {q \over {{\varepsilon _0}}}$

Important Points on Gauss’s Law

1. Gauss's law and Coulomb's law are equivalent. If we assume Coulomb's law, we can prove Gauss's law and vice versa. To prove Gauss's law from Coulomb's law, consider a hypothetical spherical surface called Gaussian surface of radius r with point charge q at its center. By Coulomb's law intensity at a point P on the surface will be $\mathop E\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^3}}}\mathop r\limits^ \to $ electric flux linked with area $\mathop {ds}\limits^ \to $ $\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^3}}}\mathop r\limits^ \to \bullet \mathop {ds}\limits^ \to $ direction of $\mathop r\limits^ \to $ and $\mathop {ds}\limits^ \to $ are same i.e., $\mathop r\limits^ \to \bullet \mathop {ds}\limits^ \to $ = r ds cos 0° = r ds. So,$\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}ds$ or $\oint\limits_s {} \mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = \oint\limits_s {} {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}ds$ for all points on the sphere r = constant $\oint {} \mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}\oint\limits_s {} ds = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}(4\pi {r^2})$ = ${q \over {{\varepsilon _0}}}$ [as$\oint {ds = 4\pi {r^2}} $] The results are true for any arbitrary surface provided the surface is closed.
2. It relates the total flux linked with a closed surface to the charge enclosed by the closed surface: a) If a closed body not enclosing any charge is placed in either uniform on non-uniform electric field total flux linked with it will be zero. b) If a closed body encloses a charge q, total flux linked with the body is independent of the shape and size of the body and position of charge inside it.

Applications of Gauss’s Law

Electric Field due to a line charge:

Gauss law is useful in calculating electric field intensity due to symmetrical charge distributions. We consider a gaussian surface which is a cylinder of radius r which encloses a line charge of length h with line charge density λ. According to Gauss law $\oint {\mathop E\limits^ \to .\mathop {ds}\limits^ \to } $=${{{q_{in}}} \over {{\varepsilon _0}}}$ $\begin{align}& \int{\overset{\to }{\mathop{E}}\,.\overset{\to }{\mathop{ds}}\,} \\ & Cylindrical \\ & surface \\ \end{align}$+$\begin{align}& \int{\overset{\to }{\mathop{E}}\,.\overset{\to }{\mathop{ds}}\,} \\ & \,circular \\ & surface \\\end{align}$+$\begin{align}& \int{\overset{\to }{\mathop{E}}\,.\overset{\to }{\mathop{ds}}\,} \\ & \,circular \\ & surface \\\end{align}$=$\frac{\lambda h}{{{\varepsilon }_{0}}}$ $\begin{align}&\int{Eds\,\,\cos \,\,0{}^\text{o}} \\& Cylindrical \\& suface \\\end{align}$+$\begin{align}& \int{Eds\,\,\cos } \\& I\,circular \\& surface \\\end{align}$$\frac{\pi }{2}$+$\begin{align}& \int{Eds\,\,\cos } \\& II\,circular \\& surface \\\end{align}$$\frac{\pi }{2}$=$\frac{\lambda h}{{{\varepsilon }_{0}}}$ E(2 πr h) =$\frac{\lambda h}{{{\varepsilon }_{0}}}$ So,   E =  $\frac{\lambda }{2\ \pi \ {{\varepsilon }_{0}}\ r}$$(E\propto \frac{1}{r})$

Electric Field due to an infinite plane thin sheet of charge:

To find electric field due to the plane sheet of charge at any point P distant r from it, choose a cylinder of area of cross-section A through the point P as the Gaussian surface. The flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder. Let surface charge density = σ [caption id="attachment_4418" align="aligncenter" width="242"] Electric Field due to an infinite plane thin sheet of charge[/caption] According to gauss law $\oint {\mathop E\limits^ \to .\mathop {dS}\limits^ \to = {q_{in}}/{\varepsilon _0}} $ $\int\limits_{\begin{smallmatrix}I\,circular \\surface\end{smallmatrix}}{E\,\,ds\cos \theta+}\int\limits_{\begin{smallmatrix}II\,\,circular \\surface\end{smallmatrix}}{E\,\,ds\cos \theta+}\int\limits_{\begin{smallmatrix}cylindrical \\surface\end{smallmatrix}}{E\,\,ds\cos \theta =}\frac{\sigma A}{{{\varepsilon }_{0}}}$ or EA + EA + 0 = ${{\sigma A} \over {2{\varepsilon _0}}}$ or E =${\sigma \over {2{\varepsilon _0}}}$

Points to remember

1. The magnitude of the electric field due to the infinite plane sheet of charge is independent of the distance from the sheet.
2. If the sheet is positively charged, the direction of the field is normal to the sheet and directed outward on both sides. But for a negatively charged sheet, the field is directed normally inwards on both sides of the sheet.

Electric Field Intensity due to uniformly charged spherical shell:

We consider a thin shell of radius R carrying a charge Q on its surface

1. At a point P₀ outside the shell (r > R) [caption id="attachment_4421" align="aligncenter" width="168"] Gauss's Law at a point outside the shell (r > R)[/caption] According to gauss law $\oint\limits_{{S_1}} {\mathop {{E_0}}\limits^ \to .\mathop {ds}\limits^ \to } $ = ${Q \over {{\varepsilon _0}}}$ or E₀ (4πr²) =${Q \over {{\varepsilon _0}}}$ E₀ = ${Q \over {4\;\pi \;{\varepsilon _0}{r^2}}}$ = ${\sigma \over {{\varepsilon _0}}}$${{{R^2}} \over {{r^2}}}$ where the surface charge density σ = ${{total\,\,ch\arg e} \over {surface\,\,area}}$ = ${Q \over {4\pi {R^2}}}$ The electric field at any point outside the shell is same as if the entire charge is concentrated at center of shell.
2. At a point P_s on surface of shell (r = R) E_S = ${Q \over {4\pi {\varepsilon _0}{R^2}}}$ = ${\sigma \over {{\varepsilon _0}}}$
3. At a point P_in inside the shell (r < R) [caption id="attachment_4424" align="aligncenter" width="427"] Gauss's Law at a point inside the shell (r < R)[/caption] According to gauss law $\oint\limits_{{S_2}} {\mathop E\limits^ \to .\mathop {ds}\limits^ \to } $ = ${{{q_{in}}} \over {{\varepsilon _0}}}$. As enclosed charge q_in = 0. So, E_in = 0. The electric field inside the spherical shell is always zero.

Electric field intensity due to a spherical uniformly charge distribution:

We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributed throughout the volume. The charge density ρ= ${{total\,\,ch\arg e} \over {total\,\,volume}}$ = ${Q \over {{4 \over 3}\pi {R^3}}}$ = ${{3Q} \over {4\pi {R^3}}}$

1. At a point P₀ outside the sphere (r > R) According to gauss law $\oint {{{\mathop E\limits^ \to }_0}.\mathop {ds}\limits^ \to } $ = ${Q \over {{\varepsilon _0}}}$ or E₀ (4πr²) = ${Q \over {{\varepsilon _0}}}$ or E₀ = ${Q \over {4\;\pi \;{\varepsilon _0}{r^2}}}$ = ${\rho \over {3\;{\varepsilon _0}}}$$\left( {{{{R^3}} \over {{r^2}}}} \right)$ [caption id="attachment_4427" align="aligncenter" width="239"] Gauss's Law at a point outside the sphere (r > R)[/caption]
2. At a point P_s on surface of sphere (r = R) E_s =${Q \over {4\pi \;{\varepsilon _0}{R^2}}}$ = ${\rho \over {3{\varepsilon _0}}}$ R
3. At a point Pin inside the sphere (r < R) According to Gauss law, [caption id="attachment_4430" align="aligncenter" width="362"] Gauss's Law at a point inside the sphere (r < R)[/caption] $\oint {{{\mathop E\limits^ \to }_{in}}.\mathop {ds}\limits^ \to } $=${{{q_{in}}} \over {{\varepsilon _0}}}$=${1 \over {{\varepsilon _0}}}$ ρ. ${4 \over 3}$πr³ = ${{Q{r^3}} \over {{\varepsilon _0}{R^3}}}$ E_in(4πr²) =${{Q{r^3}} \over {{\varepsilon _0}{R^3}}}$ or E_in = ${{Qr} \over {4\pi \;{\varepsilon _0}{R^3}}}$ = ${\rho \over {3{\varepsilon _0}}}$ r (E_in ∝ r)

Example based on Gauss’s Law