Electrostatics is the analysis of static or slow-moving electrically charged particles. Electrostatics can be seen in the following phenomena: laser machines and photocopiers, grain silos igniting instantly, and paper being attracted to a charging scale. While Gauss Law, also known as Gauss’ flux theorem of Gauss’ theorem, is the law that describes the relationship between electric charge distribution and the consequent electric field. Some key topics covered under this unit are electric charges, electric flux, electric fields, application of Gauss law.
Gauss’s law and electrostatics potential is a significant unit in the NEET 2022 Physics Syllabus, with approximately 3-4 questions anticipated yearly.
According to last year’s study, Gauss’s law and electrostatics potential questions in the NEET 2022 Physics examination will be moderately challenging. Thus, students must refer to these revision notes to help them score well in this section.
Check the entire NEET Physics Syllabus 2022 here!
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|NEET 2022 Physics: Gauss’s Law and Electrostatic Potential Notes
NEET 2022 Physics: Gauss’s Law and Electrostatic Potential Notes
Gauss law defines the proportional relationship between the total quantity of electric flux travelling through any closed path to the encased electric charge. Gauss Law is mostly used to determine the electrostatic potential due to continuous symmetries, such as:
- Uniformly charged straight wire
- Uniformly charged thin spherical shell
- Uniformly charged infinite plate sheet
Gauss Law, often known as Gauss’ theorem or Gauss’ flux theorem, is the law that explains the relationship between electric charge distribution and the consequent electric potential. The surface’s electrical field is computed using Coulomb’s law, but the Gauss law is necessary to determine the dispersion of the electric current on a closed path. It describes the static electricity confined or existing in the sealed closed path.
Gauss law refers to the total flux available within a closed surface being 1/0 times the total electric charge contained by the closed surface.
For instance, a point charge q is placed within a cube with edge’ a’. According to the gauss theorem, the flux across each surface of the cube will be “q/6πε0”.
Also See: Electric Charges and Fields Notes
Application of Gauss’s Law
Gauss’s law can be utilised effectively with anomalous symmetry to solve complex electrostatic issues, such as spherical, cylindrical, or planar symmetry. It also aids in the computation of the electrical field, which is highly complicated and requires difficult integration.
To evaluate the electrical field directly, students can use Gauss’s theorem.
Gauss law has the following major applications:
- Electric field because of uniformly charged infinite straight wire.
- Electric field because of a uniformly charged thin spherical shell.
- Electric field because of uniformly charged infinite plate sheets.
Electric Field due to Infinitely Charged Straight Wire
Imagine an infinitely long wire with a constant load density of “λ” with length “l”. Assume the Gaussian cylindrical field is related to wire symmetry when calculating the electric fields.
- The electric flow throughout the end of the cylindrical surface will be zero since the electric fields and the area vector are perpendicular to one another.
- The only electric flux that flows would be through the curved Gaussian surface.
- The magnitude of the electric field will be constant because it is perpendicular to each point of the curved surface.
- The curved cylindrical surface will have a surface area of 2πrl. The electric flow along the curve is E × 2πrl.
And as per Gauss’s law,
Φ = q / ε0
E × 2πrl = λl / ε0
E = λ / 2πε0r
Vector can define the above relationship as
→E = λ / 2πε0r n
Where “n” denotes a circular unit vector pointing toward the applied field E.
Point 1: If the linear charge density is positive, the electric field will be circumferentially outward; if the linear charge density is negative, the electric field will be circumferentially inward.
Point 2: Gaussian surface only considers the enclosed charge within it.Point
Point 3: It is critical to assume that the wire is infinitely long because otherwise, the electric current would not be perpendicular to the curved cylindrical Gaussian surface and would be at an angle with it. Check all the Important Physics Concepts for free here!
Electric Field due to Infinite Plate Sheet
Consider an infinite plane sheet with a surface charge density σ and a cross-sectional area of A. The view of the infinite plane sheet is illustrated below:
Infinite Charge Sheet
The electric field created by the infinite charge sheet would be perpendicular to the sheet’s plane. Imagine a cylindrical Gaussian surface with an axis normal to the sheet’s plane. Students can use Gauss’s law to calculate the electric field E.
Gauss’s law states that:
ϕ = q / ϵ0
Charge q will be σA as a consequence of continuous charge distribution. The gaussian surface only looks at electric flux from the two ends when discussing net electric flux. The curved surface area and an electric field are regular, resulting in zero electric flux. As a result, the net electric flux would be
Φ = EA – (– EA)
Φ = 2EA
Which can be written as,
2EA = σA / ϵ0
The term A cancels out, implying that the electric field generated by an infinite plane sheet is reflective of cross-section area A and equals
E = σ2 / ϵ0
Vector can write the above equation as
→ E = λ / 2πϵ0r n
n = direction of the current field perpendicular to and away from the infinite sheet
Point 1: If the surface charge density is positive, the electric field is directed away from the infinite sheet; if the surface charge density is negative, the electric field is directed towards the infinite sheet.
Point 2: The infinite sheet’s electric field is reflective of its position.
Electric Field due to thin Circular Shell
Imagine the radius “R” and the thin circular shell of the density of the surface electric charge. The shell exhibits circular symmetry, as may be shown from the observation. The electric field caused by the spherical shell can be also evaluated in two ways:
Electric Field Outside the Spherical Shell
Electric Field Inside the Spherical Shell
- Electric Field Outside the Spherical Shell
To determine an electric field outside the spherical shell, consider a point P outside the shell at a distance r from the shell’s centre. Gaussian spherical objects will use symmetry to create a surface with radius r and centre O.
As all points are equally spaced “r” from the centre of the sphere, the Gaussian surface will move through P and insight a consistent electric field. Then,
As per gauss’s law,
ϕ = q / ϵ0
The enclosed charge q within the Gaussian surface is σ × 4 πR2. The total electric flux perpendicular to the Gaussian surface would be
Φ = E × 4 πr2
According to Gauss’s law, it can also be written as,
E × 4πr2 = σ × 4π R3 / ϵ0
E = σR2 / ϵ0r2
Using the value of surface charge density σ as q /4πR2, students can also write the electric field as
In vector form, the electric field can be represented as
→ E = kq / r2 ŕ
ŕ = radius vector representing the electric field’s direction
Point: If the surface charge density is negative, the electric field will be directed circumferentially inward.
- Electric Field Inside the Spherical Shell
Consider point P inside the shell to calculate the electric field within the spherical shell. The spherical Gaussian surface that crosses through P is symmetrically focused at O and has a radius of r. Now, according to Gauss’s law,
ϕ = q / ϵ0
The net electric flux would be E × 4πr2. However, because surface charge density is spread outside the surface, the enclosed charge q will be 0, implying that there is no charge or current within the spherical shell. Then, according to Gauss’s Law,
E = 4
πr2 = 0
E = 0
Point: Because there is no confined charge, there is no electric current within the spherical shell.
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Gauss’s Law Formula
According to Gauss’s theorem, the total charge concealed in a closed path is directly proportional to the total flux concealed by the surface. As a result, if Φ is the total flux and ε0 is the electrically constant, the total electric charge Q concealed by the surface is as follows:
Q = Φε0
Students can write the formula for Gauss’s law as follows,
Φ = Q / ϵo
Q = total charge within the mentioned surface,
ε0 = the electric constant.
According to Gauss’s theorem, the net flux through a closed path is proportional to the net charge in the volume sealed by the closed path.
Φ = → E.d → A = qnet/ε0
In layman’s terms, Gauss’s theorem connects the ‘flow’ of electric potential lines (flux) to charges on the closed path. The net electric flux is zero if no charges are closed on the path.
It indicates that the number of electric potential accessing the surface equals the number of field lines exiting it.
Important Benefits of Gauss’s Theorem
The electrical flux from any enclosed surface is caused by the electrical fields’ sinks (negative charges) and sources (positive charges). Any electrostatics outside the surface does not affect the flow of electricity. Furthermore, only electrostatic interactions can act as electrical field sources or sinks. For instance, adjusting magnetic fields cannot act as electrical field sources or sinks.
Because the area on the left encloses a net charge, its net flux is non-zero. The net flux for the area on the right is zero because it contains no charge.
Students should note that Gauss’s law is simply a reaffirmation of Coulomb’s law. Students can easily obtain Coulomb’s law by applying the Gauss theorem to a charged object encircled by a sphere.
Preparing for the NEET 2022 Physics Gauss’ law and potential electrostatics exam is an important step toward understanding the potential of keeping energy. Students will learn the responses to questions which may have been presented in the examinations by using these revision notes. Besides being an important topic in the NEET 2022 Physics syllabus, questions on Gauss’ law and electrostatic potential carry significant points. Reading these notes will provide students with further insights on this topic which will ultimately help them in improving their scores!
1. What is Coulomb’s law?
The force between two charges is directly proportional to the product of their magnitudes and indirectly proportional to the square of their distance.
It is mathematically represented as
F ∝ | q1 | | q2 |
F ∝ 1 / r2
F = k | q1 | | q2 | / r2
where the proportionality constant is k = 1 / 4πϵ0 = 9 Х 109 Nm2C2
The permissibility of free space is ϵ0 = 8.85 Х 10-12C2N-1m-2
Coulomb’s law in vector form is →F12 = k | q1 | | q2 | / r2 r
2. What is Electric Potential?
The electric or electrostatic potential is the volume of work done to move a unit positive test current from infinity to any place along any path. It is represented by the letter V.
V = W / q0 = work done in bringing unit positive test charge from infinity to some point/unit positive test charge.
Electric potential is a vector value with volt as the SI unit.
The electric potential related to a point charge q at separation r is equal to V = 1 / 4πϵ X q / r
Students will answer electric potential due to several charges q1, q2,… qn.
= 1 / 4πϵ X ( q1/r1 + q2/r2 + … + qn/rn)
= 1 / 4πϵ ∑ qi / ri
Electric Field and Potential Relationship
Students can describe the relation between electric field and potential as
E = -dV / dr
E represents the electric field.
V represents the potential difference.
3. Define conductors, semiconductors or insulators.
A conductor is a body through which an electric charge can easily pass. Metals are good examples of conductors.
Insulators, often known as dielectrics, are materials that do not conduct electricity. Rubber, glass, plastics, wool, and other materials are examples of insulators.
Semiconductors are the bodies that exist between conductors and insulators. Examples include silicon, germanium, and others.
4. Define electric lines of force?
An electric line of force is a straight or curved path through which a positive unit charge is compelled to go when free to do that in an electric field. The velocity of a unit positive charge determines the direction of the line of force.
The force lines diverge from a positive charge conductor toward a negative charge conductor.
A forced line begins with a positive charge and finishes with a negative charge. It indicates that the line of force begins at a greater potential and ends at a lower potential.
5. What are the important pointers of Gauss’s law?
The primary application of Gauss law is to determine the electric field produced by:
- A uniformly charged infinite straight wire.
- An endless plate sheet with a uniform charge.
- A thin uniformly charged spherical shell.
- Gauss’s law explains the electrical charge confined in the enclosed or electrical charge in the enclosed closed surface.
- Gauss Law states that the total flux associated with a closed path is 1/0 times the charge encompassed by the closed surface.
- The Gauss law formula is written as Φ = Q/ε0. Q is the total charge on the given surface, and 0 is the electric constant.
- Students can utilise Gauss’ law to address complex electrostatic issues with unique symmetry such as spherical, cylindrical, or planar.