Gauss Law - Applications, Gauss Theorem Formula

Gauss’s Law states that the electric flux Φ through a closed surface is equal to the charge Q enclosed divided by the permittivity ϵ of the medium: $$\Phi = \frac{Q}{\epsilon}$$ The electric flux is defined as the electric field E multiplied by the area A of the surface projected in a plane and perpendicular to the field: $$\Phi = EA$$

Table of Contents:

• What is Gauss Law?
• Gauss’s Law Formula
• The Gauss Theorem
• Applications of Gauss Law
• Frequently Asked Questions on Gauss’ Law
• Problems on Gauss Law

What is Gauss Law?

Gauss Law states that the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

According to Gauss’s Law, the total flux linked with a closed surface is equal to $$\frac{1}{\varepsilon_0}$$ multiplied by the charge enclosed by the closed surface.

$\oint \overrightarrow{E}.\overrightarrow{d}s=\frac{1}{{\epsilon }_0}q$

As per Gauss law, when a point charge q is placed inside a cube of edge ‘a’, the flux through each face of the cube is q/6ε0.

The electric field is a fundamental concept to understanding electricity. Generally, the electric field of a surface is calculated using Coulomb’s law. However, to calculate the electric field distribution in a closed surface, one must understand the concept of Gauss law. This law explains the electric charge enclosed in a closed or the electric charge present in the enclosed closed surface.

Gauss Law Formula

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Therefore, if ϕ is the total flux and ${ϵ}_0$ is the electric constant, the total electric charge Q enclosed by the surface is given by:

$$Q = \epsilon_0 \phi$$

as per Gauss’ theorem.

$Q=\varphi {ϵ}_0$

The Gauss law formula is expressed as:

$Q/{ϵ}_0=\varphi$

Q = Total charge within the given surface

${ϵ}_0$ = the electric constant.

⇒ Further Reading: Equipotential Surface

The Gauss Theorem

The net flux through a closed surface is proportionate to the net charge contained within the closed surface.

$\Phi =\to E\cdot d\to A=\frac{q}{{ϵ}_0}$

In simple words, the Gauss’s Law states that the total electric flux through a closed surface is equal to the charge enclosed by the surface. If no charges are enclosed by a surface, then the net electric flux is zero.

The number of electric field lines entering the surface is equal to the number of field lines leaving the surface.

An important corollary of the Gauss theorem is that:

The electric flux from any closed surface is only due to the sources (positive charges) and sinks (negative charges) of the electric fields enclosed by the surface. Any charges outside the surface do not contribute to the electric flux. Also, only electric charges can act as sources or sinks of electric fields. Changing magnetic fields, for example, cannot act as sources or sinks of electric fields.

Gauss Law in Magnetism

The net flux for the surface on the left is non-zero as it encloses a net charge. The net flux for the surface on the right is zero since it does not enclose any charge..

Note: The Coulomb’s law can be restated using the Gauss law. If the Gauss theorem is applied to a point charge enclosed in a sphere, it is easy to see that the result is the same as Coulomb’s law.

Applications of Gauss’s Law

1. At the centre, x = 0 and thus, the electric field, $E=\frac{1}{4\pi \in \mathrm{\_}0}\frac{qx}{{(R^2+x^2)}^{3/2}}$ is equal to 0.

2. The electric field, E, at a distance r from an infinite line of charge is given by $$E = \frac{\lambda}{2\pi r \epsilon_0} = \frac{1}{4\pi \epsilon_0} \cdot \frac{2\pi}{r}$$ where λ is the linear charge density.

3. The electric field intensity near a plane sheet of charge is given by $$E = \frac{\sigma}{2\epsilon_0K}$$, where σ is the surface charge density.

4. The intensity of the electric field near a plane charged conductor is given by $E=\frac{\sigma }{K{ϵ}_0}$ in a medium of dielectric constant K. If the dielectric medium is air, then $E_{air}=\frac{\sigma }{{ϵ}_0}$.

5. The field between two parallel plates of a condenser is $$E = \frac{\sigma}{\epsilon_0}$$, where σ is the surface charge density.

Gauss Law and its Application to an Infinite Wire’s Electric Field

The electric field due to an infinitely long line of charge with charge per unit length λ has cylindrical symmetry, with all electric fields pointing radially away from the line of charge and no component parallel to the line of charge.

We can use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface, which has a radius (r) and length (l).

Applications of Gauss Law – Electric Field due to Infinite Wire

The angle between the electric field and area vector is zero, and thus cosθ = 1, as can be seen in the diagram above.

The top and bottom surfaces of the cylinder are parallel to the electric field. Thus, the angle between the area vector and the electric field is 90 degrees, and cosθ = 0.

Therefore, the electric flux is caused only by the curved surface.

According to Gauss’s Law,

$\Phi =\to E.d\to A$

$\Phi ={\Phi }_{curved}+{\Phi }_{top}+{\Phi }_{bottom}$

$\Phi =\int \overrightarrow{E}\cdot d\overrightarrow{A}\cos 0+\int \overrightarrow{E}\cdot d\overrightarrow{A}\cos {90}^{\circ }+\int \overrightarrow{E}\cdot d\overrightarrow{A}\cos {90}^{\circ }$

$\Phi =\int E·dA·1$

The electric field on the surface of an object with radial symmetry is constant in magnitude due to the fact that it is equidistant from the line of charge.

$\Phi =2\pi rE\int dA$

The net charge enclosed by the surface is:

$Qnet=\lambda .l$

Using Gauss’ Theorem

$\Phi =E\times 2\pi rl=q_{net}/{\epsilon }_0=\lambda l/{\epsilon }_0$

$E\times \frac{2\pi rl}{\lambda l/{ϵ}_0}=1$

$E=\frac{\lambda }{2\pi r{ϵ}_0}$

⇒ Further Reading: Electric Potential Energy

Problems on Gauss’s Law

Problem 1: Using the Gauss theorem, calculate the flux of a uniform electric field of magnitude E = 100 N/C through a plane square area of edge 10 cm placed in the Y-Z plane, with the normal along the positive X-axis being positive.

Solution:

$\Phi =\int E.cos\theta ds$

The angle θ is equal to 0 in the direction of the electric field normal to the area.

$\Phi =E.\Delta S=(100N/C)(0.10m{)}^2=1N-m^2$

Problem 2: Find the flux of the electric field through a circular area of radius 1 cm lying in the region where x, y, and z are all positive, and with its normal making an angle of 60° with the Z-axis, on a large plane charge sheet having surface charge density σ = 2.0 × 10⁻⁶ C-m⁻².

Solution:

The electric field near the plane charge sheet is $E=\sigma /2\epsilon 0$ in the direction away from the sheet. At the given area, the field is along the Z-axis.

The area = $\pi r^2=3.14\times 1{\text{ cm}}^2=3.14\times {10}^{-4}{\text{ m}}^2$.

The angle between the normal to the area and the field is 60°.

Hence, according to Gauss’s theorem, the flux $\overrightarrow{E}.\mathrm{\Delta }\overrightarrow{S}$

$E.\Delta Scos\theta$

$\sigma /(2\epsilon ₀\ast r²\ast cos(60°))$

$\frac{2.0\times {10}^{-6}C}{2\times 8.85\times {10}^{-12}{C}^2}\times \frac{3.14\times {10}^{-4}{m}^2}{N-m^2}\times \frac{1}{2}=17.5N-m^2{C}^{-1}$

Problem 3: A charge of 4×10^-8C is distributed uniformly on the surface of a sphere with a radius of 1 cm. This sphere is covered by a concentric, hollow conducting sphere with a radius of 5 cm.

Find the electric field at a point 2 cm away from the center.

Find the surface charge density on the outer surface of the hollow sphere given that a charge of 6 x 10⁻⁸C is placed on the hollow sphere.

Solution:

(a) Consider Figure (i).

Draw a concentric spherical surface through point P. All points on this surface are equivalent; thus, the magnitude and radial direction of the field at all these points will be equal.

The flux through this surface = $\oint \overrightarrow{E}\cdot \overrightarrow{dS}$

$\oint EdS=E\oint dS$

$4\pi x^{2}E$

x = 2 cm = 2 × 10^-2 m.

Thus, from Gauss’ Law, the flux is equal to $\frac{q}{{ϵ}_0}$.

$E=\frac{q}{4\pi {ϵ}_0{x}^2}$

$9\times {10}^{1}4\times [(4\times {10}^{-8})/(4\times {10}^{-4})]=9\times {10}^{5}N{C}^{-1}.$

(b) Let’s take a look at Figure (ii).

The flux $\oint \overrightarrow{E}.d\overrightarrow{S}$ through the Gaussian surface taken through the material of the hollow sphere is zero, since the electric field in a conducting material is zero.

The charge on the inner surface of the hollow sphere is 4 × 10^-8C, as per Gauss’ Law, the total charge enclosed must be zero.

The charge on the inner surface will be 6 × 10^-8 C, and the charge on the outer surface will be 10 × 10^-8C.

Problem 4: Find the charges appearing on the surfaces of shells B and C given that three concentric thin spherical shells A, B, and C of radii a, b, and c, respectively, with shell A and C given charges q and -q respectively and shell B earthed.

Solution:

Suppose the inner surface of B has a charge -q and the outer surface of B has a charge q’, as shown in the previous worked out example from Gauss law.

The inner surface of C must have a charge of -q, as per Gauss law. This means that the outer surface of C must have a charge of q’ - q, as the net charge on C must be -q. The charge distribution is illustrated in the figure below.

The Potential at B

The charge q on A is equal to $q/(4\pi \epsilon ₀b)$

The charge on the inner surface of B is equal to $-q/4\pi \epsilon 0b$

The charge q’ on the outer surface of B is equal to $q^{\prime }/4\pi \epsilon 0b$

Due to the charge -q’, the electric field on the inner surface of C is equal to $-q^{\prime }/4\pi {\epsilon }_{0}c$

The charge on the outer surface of C is $(q^{\prime }-q)/4\pi \epsilon ₀c.$

The net potential is,$V_B=\frac{q^{\prime }}{4\pi {ϵ}_{0}b}-\frac{q}{4\pi {ϵ}_{0}c}$

Thus, q’ = 0

The charges on various surfaces are illustrated in the figure below:

Problem 5: What charge should be given to a particle of mass 5 × 10^-6g so that if released, it does not fall down when kept over a large horizontal sheet of charge density 4.0 × 10^-6C/m²(figure)? How many electrons are to be removed to give this charge? How much mass is decreased due to the removal of these electrons?

Solution:

The electric field in front of the sheet is:

E = 2.26 × 10⁵ N/C

The electric force qE acting in the upward direction will balance the weight of the particle if a charge q is given to the particle.

q × 2.26 × 10⁵ N/C = 5 × 10^-9 kg × 9.8 m/s²

$q=[2.21\times {10}^{-13}C]/[2.26\times {10}^{5}C]=4.9\times {10}^{-8}$

The number of electrons to be removed to create a charge of 1.6 × 10^-19C is 1.

= 1.4 × 10⁶

Mass decreased due to the removal of these electrons = 1.3 x 10^-24 kg

Problem 6: Find the distribution of charges on the four surfaces of two parallel conducting plates, A and B, given that A has a charge of Q1 and B has a charge of Q2.

Solution:

Consider a Gaussian surface as shown in Figure (a). Two faces of this closed surface are located completely inside the conductor, where the electric field is zero.

The flux through these faces is zero. The other parts of the closed surface, which are outside the conductor, are parallel to the electric field, and hence the flux on these parts is also zero.

From Gauss law, the total flux of the electric field through the closed surface is zero, thus implying that the total charge inside the closed surface should also be zero. Furthermore, the charge on the inner surface of A should be equal and opposite to that on the inner surface of B.

![Gauss Law Questions - 6B]()

The distribution should be like the one shown in Figure (b). In order to find the value of q, one should consider the field at a point P inside Plate A. Additionally, the surface area of the Plate (one side) should be taken into account.

The electric field at P can be calculated using the equation $E=\sigma /2{\epsilon }_0$.

Due to the charge $Q1–q=\frac{(Q1–q)}{2A{ϵ}_0}$ (downward)

The charge has a resulting electric field of $\frac{q}{2A{ϵ}_0}$ (upward)

The charge -q has a magnitude of $q/2A\epsilon 0$, directed downward.

Due to the charge, $Q_2+q=(Q_2+q)/2A{\epsilon }_0$ (upward).

The net electric field at P (in the downward direction) due to all the four charged surfaces is

$-(Q_1-q)/2A{ϵ}_0+q/2A{ϵ}_0-(Q_2+q)/2A{ϵ}_0$

As the point P is inside the conductor, this field should be zero.

Hence, $Q_1-q+Q_2-q=0$

$Q=\frac{Q_1-Q_2}{2}$ (i)

Thus, $$Q_1 - q = \frac{Q_1 + Q_2}{2} \ldots \ldots (ii)$$

$$Q_2 + Q_1 = \frac{[Q_1 + Q_2]^2}{2}$$

The distribution shown in figures (a, b) can be redrawn using these equations, as seen in the figure.

The two outermost surfaces of charged conducting plates placed parallel to each other have equal charges, and the facing surfaces have equal and opposite charges; this result is a special case of the aforementioned result.

Problem 7: What will be the potential difference between the surface of a solid conducting sphere with charge Q and the outer surface of a concentric, uncharged conducting hollow spherical shell if the shell is given a charge of -3Q?

Solution:

In Case of a Charged Conducting Sphere

$Vin=Vc=Vs=\frac{1}{4\pi {ϵ}_0}$

V_out = $\frac{1}{4\pi {ϵ}_0}$

The potential at the surface of a sphere with radius a and a spherical shell with radius b will be;

$V_{sphere}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{a}and{V}_{shell}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{b}$ and so, according to the given problem;

$\frac{Q}{4\pi {ϵ}_0}[\frac{1}{a}-\frac{1}{b}]=V\dots \dots \dots \dots \dots \dots \dots \dots (1)$

The potential at the surface and inside of the shell when given a charge of -3Q will change.

$V_0=\frac{1}{4\pi {\epsilon }_0}[-\frac{3Q}{b}]$

So, now

$V^{\prime }\text{sphere}=\frac{1}{4\pi {ϵ}_0}[\frac{Q}{a}+V_0]\text{and}{V}^{\prime }\text{shell}=\frac{1}{4\pi {ϵ}_0}[\frac{Q}{b}+V_0]$

Hence, $$V_{sphere} - V_{shell} = \frac{Q}{4\pi\epsilon_0} \left[\frac{1}{a} - \frac{1}{b}\right] = V \quad \text{[from Eqn. (1)]}$$

The potential difference between the sphere and the shell will remain unchanged, even if a charge is applied to the external shell.

The potential difference between sphere and shell will remain unchanged due to the presence of charge on the outer shell, causing the potential everywhere inside and on the surface of the shell to change by the same amount.

Problem 8: Calculate the charge q of a very small sphere of mass 80 g with an equal charge q, which is held at a height of 9 m vertically above the centre of a fixed nonconducting sphere of radius 1 m. When released, the sphere falls until it is repelled just before it comes in contact with the sphere. [g = 9.8 m/s2]

Solution:

Keeping in mind that here both electric potential energy and gravitational potential energy are changing, and for an external point, a charged sphere behaves as if the whole of its charge were concentrated at its center.

Applying the law of conservation of energy between the initial and final positions, we can conclude that

$(1/4\pi \epsilon 0)\times (q^2/1)+mg\times 1=(1/4\pi \epsilon 0)\times (q^2/9)+mg\times 9$

$q^2=(80\times {10}^{-3}\times 9.8)/109=28\mu C.$

Solved Questions on Gauss’s Law

What is the Relationship between Gauss Law and Coulomb’s Law?

We can say that Gauss’s law and Coulomb’s law are equivalent, meaning they are essentially the same. This relationship can be demonstrated through a derivation with the help of an example, which is discussed extensively in electrodynamics.

If we apply Coulomb’s law to a point charge q, the electric field generated is given by:

E = kq/r²

If we take a sphere of radius (r) centred on charge q, then the surface S of this sphere will have an electric field given by $$E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$

At the end of the equation, we can see that it refers to Gauss law. By deducing the spherical symmetry of the electric field and performing integration, we can determine the relation between Gauss law and Coulomb’s law. All in all, it is clear that both the laws are related to each other.

What Factors Should We Consider When Choosing an Appropriate Gaussian Surface?

In order to choose an appropriate Gaussian Surface, we have to take into account the ratio of charge and the dielectric constant which is given by a two-dimensional surface integral of the electric field symmetry of the charge distribution. There are three different cases that we need to be aware of.

When the charge distribution is spherically symmetric, it is referred to as “spherical”.

When the charge distribution is cylindrically symmetric, it is referred to as cylindrical.

When the charge distribution has translational symmetry along a plane, it is referred to as Pillbox.

We can select the size of the surface based on where we need to calculate the field. Gauss theorem is beneficial in determining a field when there is a specific symmetry, as it lets us know how the field is oriented.

What is the Relationship between Electric Flux and Gauss’s Law?

The Gauss Law states that the net electric flux through a closed surface is equal to zero if the volume defined by the surface contains no net charge.

We will begin by examining Gauss’s Law in order to establish the relation.

Gauss’s law can be represented as:

${\Phi }_E=Q/\epsilon o$

Here,

${\Phi }_E$ = Electric flux through a closed surface S enclosing any volume V.

Q = Total charge enclosed within V

$\epsilon ₒ$ = Electric Constant

The electric flux ΦE can now be defined as the surface integral of the electric field, given by:

**$\Phi E=\int \int E·dA$**s a sentence.’

Here’s a sentence.

E = electric field.

dA = vector representing an infinitesimal element of area on the surface.

Gauss’s law is known as the integral form, and it is considered to be an integral of the electric field. Notably, flux is related to this form.

How is the differential form of the Gauss Theorem expressed?

The differential form of Gauss law states that the divergence of the electric field (E) is equal to the volume charge density (p) at a particular point in space. Mathematically, it is represented as:

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

$\Delta E=\frac{\rho }{{ϵ}_o}$

ε_o = Permittivity of Free Space.

Using Gauss Law to Find the Electric Field

The Steps Involved in Solving the Electric Field with Gauss Law Are: 1.2.3.4.5.

1. Identify the spatial symmetry of the charge distribution.

2. Choose a Gaussian surface with the same symmetry as the charge distribution.

3. Identify the consequences of this choice. Calculate the integral of $$\Phi_sE$$ over the Gaussian surface, and then determine the flux through the surface.

4. Calculate the electric flux through the Gaussian surface.

5. Determine the electric field of the charge distribution.

Students must remember the three types of symmetry in order to calculate the electric field:

1. Reflectional Symmetry
2. Rotational Symmetry
3. Translational Symmetry

Spherical Symmetry

Cylindrical Symmetry

Planar Symmetry

Calculations of inappropriate coordinate systems and the corresponding correct Gaussian surface for the particular symmetry should be performed.

Frequently Asked Questions on Gauss’s Law

Can Gauss law be applied to all surfaces?

Yes, Gauss’ Law can be applied to all surfaces. For any closed surface and any distribution of charges, Gauss’s Law holds true.

Can Gauss law be applied to the non-uniform electric field?

Yes, Gauss’s Law can be applied to non-uniform electric fields. Gauss’s law can be applied to both uniform and non-uniform electric fields.

State Gauss’s law.

Gauss’s Law states that the net electric flux through any closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

According to Gauss’s Law, the net flux of an electric field through a closed surface is proportional to the charge enclosed within it.

What is the factor on which the electric field lines depend?

The electric field lines depend on the charge distribution.

Is the Flux Through the Surface Positive or Negative?

Flux through the surface is taken as positive if the flux lines are directed outwards or negative if the flux is directed inwards.

A Gaussian Surface is a closed surface in three dimensions that is used to calculate the flux of a vector field through the surface.

The electric flux is calculated through a Gaussian surface.

Yes, Coulomb’s law can be derived using Gauss’ law.

Yes, Coulomb’s law can be derived using Gauss law and vice-versa.

Surface charge density is a measure of the total electric charge per unit area of a surface. It is expressed as the amount of charge per unit area and is denoted by the Greek letter σ (sigma).

When the charge is uniformly spread over the surface of a conductor, it is referred to as surface charge density.