electric flux
or what is electric flux
or define electric flux
or electric flux definition
or electric flux class 12
or electric flux density
In easy language
The electric flux, nothing but passing the electric field lines throughout the surface is called electric flux .
Electric flux can be represented as 𝛟E And it is tell about how many electric field lines passing through unit area.
Or in different words
Electric flux, a cornerstone concept in electromagnetism, serves as the yardstick for gauging the flow of the electric field through a designated surface. This intricate concept finds its mathematical representation through the dot product of the electric field (E) and the area (A) vector, all multiplied by the cosine of the angle (θ) between the electric field vector and the surface normal. Let's delve into the formula that encapsulates electric flux:
electric flux formula
[ฺ𝛟E = E.A ]
- Electric flux is a dot product of electric field vector and area vector.
- If we talked about area vector so it is differentiate by
- For plain body
- For curved body
- For plain body
For a plain body, the area vector is always perpendicular to the surface .
2. For curved body
In this case the area vector may not be perpendicular to the surface so when it is not perpendicular to the surface we take the perpendicular component of the area vector to the surface of the body.
So :->
[ฺ𝛟E = EAcosθ ]
Decoding the Formula Components
Breaking down the formula reveals the significance of each variable:
- (𝛟E) stands for electric flux.
- (E) represents the electric field vector (electric field in space).
- (A) symbolizes the area vector (perpendicular to the surface).
- (θ) stands for the angle between the electric field vector and the normal to the surface (perpendicular area vector).
This formula succinctly conveys that electric flux is determined by multiplying the strength of the electric field, the surface area, and the cosine of the angle between the electric field vector and the surface's normal. It encapsulates the essence of how these factors intertwine to quantify the flow of the electric field through a given surface.
Q. Let's find the electric flux throughout the surface.
Electric field = 5N/c
Length and breath of plane surface 3m and 2m respectively .
So
Due to [ฺ𝛟E = E.A ]
[ฺ𝛟E = EAcosθ ]
𝛟E = 5 * (3*2) cos0
𝛟E = 30*1
𝛟E = 30
So it is telling that 30 flux passes through one unit area of the plane surface.
Simplifying the Concept
In simpler terms, electric flux can be envisioned as the tally of electric field lines traversing a specific surface. When the surface is aligned parallel to the electric field lines, the flux reaches its max. Conversely, if the surface is perpendicular, the flux registers at zero.Application in Class 12 Physics
In the realm of Class 12 physics, electric flux assumes a pivotal role within the domain of electrostatics. Its introduction facilitates a comprehensive understanding of how electric fields interact with surfaces and furnishes a quantifiable method for assessing these interactions.
Electric Flux Density: A Related Notion
Electric flux density (D), akin to electric flux, finds its relevance primarily in the realm of dielectric materials. The relationship between the two is elucidated through the formula:
[D = E/ε ]
Unveiling the Components
The key players in this formula include:
- (D), denoting electric flux density.
- (E), representing the electric field vector.
- ( ε ), signifying the permittivity of the material.
Significance in Problem Solving
Comprehending electric flux and its counterpart, electric flux density, proves to be indispensable when tackling issues related to electric fields and their interactions with diverse materials. The ability to decipher these concepts enhances problem-solving capabilities, providing a nuanced insight into the intricate dance between electric fields and the materials they encounter. As we navigate the landscape of electromagnetism, the understanding of electric flux emerges as a beacon guiding physicists through the intricate web of electric interactions.
Electric flux through inclined plane:
in case of inclined plane when we take perpendicular area vector to the surface it will make angle between electric field and area vector (90-θ) so cosθ will change into sinθ:𝛟E = EAcos(90-θ)
𝛟E = EAsinθ { by trigonometric equation }
So now
- electric flux is max when angle = 90
- And electric flux is min when angle = 0
- So while calculating the flux we take the sin component of the area vector .
So as we can easily see that electric flux passing through plane area it is same as flux passing through inclined plane it means:
𝛟E = passing through plane = passing through inclined plane.
It is called the concept of projected area.
learn more :
electric field and application (in easy way )
electric force most important concepts with examples
charge density (linear, area, volume with most helpful important examples )
in-depth about charge with experiments
what is electric flux: know every things about it with iits examples
So let’s learn more application of it.
In this case we have angle θ, length of inclined plane L and height l so let’s find electric flux passing through inclined plane?
As we can see in the figure the area vector of the inclined plane is perpendicular to the surface and the electric field is not perpendicular to the surface .
so due inclined plane makes angle θ to the surface
The angle between inclined plane and electric field vector is also θ { due to help of alternate angle} so it is clear that angle between area vector and electric field (90-θ).
now we will take help of number of electric field file:
𝛟E = E(L*l)sinθ
Or 𝛟E = E * Llsinθ
So
𝛟E = E * Ll
Let’s try this one:
Let's take a hemisphere and pass some electric field lines and try to find out electric flux.
As we see that flux passing through the plane surface of hemispherical surface = flux through curved surface area.
So 𝛟E = E * A(plane)* cos0
{due to in this case we didn’t take any component of area so we take cosθ and θ = 0 in figure}
𝛟E = E * 𝞹r^(2) *1
𝛟E = E * 𝞹r^2
below image represents flux passing through sphare.
si unit of electric flux
or unit of electric flux
or dimension of electric flux
or electric flux si unit
or dimensional formula of electric flux
or electric flux dimensional formula
The SI unit assigned to electric flux is the volt-meter (V·m), and its dimensional formula is denoted as [ML³T⁻³A⁻¹]. Here's a breakdown of the dimensional components:
- M stands for mass,
- L represents length,
- T corresponds to time,
- A signifies electric current.
Consequently, the dimensional formula for electric flux (𝛟E ) is precisely represented as [ML³T⁻³A⁻¹]. This formulation provides a comprehensive understanding of the fundamental dimensions involved in the measurement of electric flux, further enhancing its applicability in the context of electromagnetic phenomena.
electric flux is scalar or vector
Electric flux distinguishes itself as a scalar quantity within the realm of electromagnetism. Unlike its counterpart, the electric field (E), which assumes a vector nature, electric flux (𝛟E) operates as a scalar. This characteristic denotes that electric flux conveys solely magnitude without a directional component. In contrast, the electric field not only possesses magnitude but also encapsulates directionality, solidifying its classification as a vector quantity. The scalar essence of electric flux simplifies its representation, focusing solely on the quantity of electric field permeating a designated surface, unencumbered by directional intricacies.
Difference between plane and closed surface
- It is important due to it being used in gauss’s law so it is important to know what is plane and closed surface.
- In case of a closed surface area vector always outside of the surface and it is perpendicular to the surface (not always).
- In case of closed surface 𝛟E (in) and 𝛟E (out) is defined but in case of plane or not enclosed surface is not defined.
- In the case of a plane, the area vector is perpendicular to the surface. It may be positive or negative as per direction.
Electric field due to Ring:
Assume that the electric field for ring radius R at distance of x is E and angle subtended by the upper end of the ring to the line of joining alpha and from lower end beta so as we read in the case of an electric field it has two components of electric field parallel and perpendicular.
So E丄 = 2Ecos(θ/2) = E {when two E are same then net E = 2Ecos(θ/2) }
By figure
E = kQ/ (R^2 + x^2)
cosθ/2 = x/ (R^2 +x^2)
So E = 2Ecos(θ/2)
E = 2* [kQ/ (R^2 + x^2) ]* [x/ R^2 +x^2]
E = k(2Q)x / (R^2 + x^2)^3/2
Some important application of it
Electric field for any regular polygon
As we can see in figure
If equal charge in placed at the vertex of polygon then electric field:
E = k (nQ)x / (R^2 + x^2)^3/2
Where n = number of charges placed on vertex
X = distance of point from polygon
R = distance between vertex and center of polygon
So in this case the electric field at distance x by four charged particles is aligned as a vertex of square.
Find and mail me an answer.
Note: in case of triangular polygon max electric field :
E(max) = R/√2
gauss law
or gauss law class 12
or state gauss law
or gauss law in electrostatics
or gauss law of electrostatics
or gauss law and its application
or gauss law definition
or define gauss law
or gauss law statement
or state and explain gauss law
Unlocking the Mysteries of Gauss's Law in Electrostatics
Gauss's Law, a cornerstone in the realm of electrostatics, stands as a fundamental principle intertwining the electric flux through a closed surface with the net electric charge enclosed within that boundary. Nestled among Maxwell's equations, this law plays a pivotal role in unraveling the intricate distribution of electric charges.
Statement of Gauss's Law:
At its core, Gauss's Law articulates that the electric flux (𝛟E) coursing through a closed surface equals the product of the electric field (E) and the area (A) of that surface. Furthermore, it equates to the net electric charge (Q) enclosed by the surface divided by the electric constant (varepsilon_0):
[ 𝛟E = Q/ε]
Expressed mathematically as a closed surface integral (S):
∮E .dA = Q/ε
Breaking down the components:
- (∮) denotes the closed surface integral,
- (E) represents the electric field,
- (dA) symbolizes an infinitesimal area vector on the closed surface,
- (Q) embodies the net electric charge enclosed by the surface,
- (ε) stands for the electric constant or permittivity of free space.
Explanation:
The essence of Gauss's Law lies in its proclamation that the total electric flux through a closed surface harmoniously correlates with the total charge enclosed by that boundary. This principle bestows a convenient methodology for computing electric fields, particularly in scenarios endowed with high symmetry, thereby rendering problem-solving more accessible.
Applications:
Gauss's Law finds myriad applications, especially in scenarios involving symmetric charge distributions like spheres, cylinders, and planes. Its prowess lies in simplifying calculations by leveraging the inherent symmetry of the system. This law emerges as a robust tool, facilitating a profound comprehension of electric fields and their predictions across diverse situations.
gauss law in magnetism
or gauss law of magnetism
or state and explain gauss law in magnetism
In easy language
Gauss’s law states that net flux passing through an imaginary closed surface (gaussian surface ) is equal to 1/ ε times charged inclosed by the surface.
So as we can see in the figure net flux passing through closed surface is Qin closed divide by ε. where Qinclosed represents that number of charges inside the gaussian surface.
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Gauss's Law for electrostatics is eloquently expressed through the mathematical formulation:
∮E.dA = Qinclose/ε
Breaking down the components:
- \(∮) symbolizes the closed surface integral,
- \(E) represents the electric field vector,
- \(dA) embodies an infinitesimal area vector on the closed surface,
- \(Qinclose) denotes the net electric charge enclosed by the surface,
- \(ε) stands for the electric constant or permittivity of free space.
In this examples we can see that Electric field due to outer charge on the closed surface is
Qin = Quot so the amount of charge interesting and amount of charge coming through the closed surface by external charge is the same .
That ‘s why net flux passes through a closed surface due to outer charge zero.
𝛟net = 0
Note: if we put any charge outside there is no effect on electric flux due to it.
Derivation/Proof:
Consider a closed surface (S) enveloping a volume (V). According to Gauss's Law, the electric flux through (S) equates to the total charge enclosed divided by the electric constant:
∮E . dA = Qinclose/ε
Now, employing the divergence theorem, linking a surface integral to a volume integral, we express the left side using the electric field divergence
∮E . dA = ∫∫v (▽.E)dv = Qinclose/ε
By applying the divergence theorem, this transforms into:
∫∫v (▽.E)dv = Qinclose/ε
As this equation holds for any closed surface (S) and its enclosed volume (V), the integrands must be equal at every point in space:
▽.E = ρ /ε
Breaking it down:
- (▽.E) represents the divergence of the electric field,
- (ρ) signifies the charge density.
This differential form of Gauss's Law posits that the divergence of the electric field is proportional to the charge density at any point in space, providing a nuanced understanding of the intricate relationship between electric fields and charge distributions.
NOte:
Flux remains the same if the location of charge is changed into the gaussian part; it means that flux didn’t change if we moved charge into the gaussian surface .
Flux is independent of shape and size of surface .
If we put any closed surface into any uniform electric field then net flux passing throughout the surface is zero.
So when net flux passing through the surface is zero it is not necessary that the electric field passing through the surface also be zero.
application of gauss law
or applications of gauss lawor gauss law application
1. Flux due to any point charge
In case of point charge flux passing through any radial close surface is
E . dA = Qinclose/ε = 𝛟net
If we talk about 𝛟net what is really means of it , so it means that number of electric field lines passing through closed surface .
2. Distribution of flux
It is the most important part in electric flux so let’s learn more important examples of it .
In case of plane (on the surface) :
When charge placed exposed upper part of plane then number of electric field passing through that surface is half of electric field.
𝛟net = ( Q/ ε)*1/2
𝛟net = ( Q/ 2ε)
In case of square (on the surface)
So when charge pleased on the surface of square then number of electric field passing through the surface is half of electric flux and remaining electric field not passing through surface .
So 𝛟net = ( Q/ 2ε)
So it is same as plane surface.
In case of square (when charge played on the center of it)
So in this case when charge placed on the center of square then there are 6 surfaces through which electric filed passes so , number of electric filed passes through all surfaces is
𝛟net = ( Q/ ε)
So number of electric field passing through one surface is
𝛟net = ( Q/ ε)*1/6 (due to 6 surfaces by which electric field passes.)
NOTE: similarly we can find number of electric field or flux passing through all surfaces and as well as from one surface also.
3. Uniformly Charged Insulating Sphere:
- Gauss's Law finds application in determining the electric field inside and outside a uniformly charged insulating sphere, serving as a fundamental example in introductory physics.
4. Electrostatic Shielding:
- Understanding the principles of Gauss's Law aids in predicting and mitigating electric fields inside conductors. In electrostatic equilibrium, the electric field within a conductor is null, forming the basis for electrostatic shielding techniques to protect sensitive equipment.
5. Dielectric Materials:
- Gauss's Law is a pivotal tool for analyzing the electric field within dielectric materials. Its application enhances comprehension of how dielectrics influence the electric field in devices like capacitors.
6. Charged Conductive Shells:
- Gauss's Law facilitates the determination of electric fields inside and outside charged conducting shells, revealing a critical insight that the electric field within the shell is consistently zero, irrespective of the charge distribution.
7. Electrostatic Induction:
- Gauss's Law plays a key role in explaining electrostatic induction, elucidating the redistribution of charges on conductive objects influenced by nearby charged entities.
8. Verification of Coulomb's Law:
- Gauss's Law provides an alternative avenue for verifying Coulomb's Law by demonstrating the equivalence of the electric field outside a charged sphere to that produced by a point charge.
9. Flux Capacitors and Electrostatic Imaging:
- In practical applications, Gauss's Law is indispensable for designing and analyzing electrostatic devices such as flux capacitors and electrostatic imaging systems.
10. Understanding Electric Fields in Various Geometries:
- Gauss's Law emerges as a powerful ally in understanding and calculating electric fields within intricate charge distributions and geometries. Its utility shines brightest in scenarios marked by high symmetry, simplifying the problem-solving landscape.
These diverse applications underscore the versatility and indispensability of Gauss's Law in deciphering and forecasting electric fields across a spectrum of electrostatic scenarios.
deduce coulomb's law from gauss law
or derive coulomb's law from gauss lawDeriving Coulomb's Law from Gauss's Law: A Step-by-Step Journey
Assumption:
Let's consider a spherically symmetric charge distribution with a total charge (Q) concentrated at the center of a sphere with radius (r).
Steps:
Choose a Gaussian Surface:
- Begin by selecting a Gaussian surface, and for our spherically symmetric scenario, a spherical surface centered on the point charge is apt.
Apply Gauss's Law:
- Employ Gauss's Law to the chosen surface:
E . ∮dA = Qinclose/ε
Simplify and Exploit Symmetry:
- Exploit the spherical symmetry, rendering the electric field (E) radial. The angle between (E) and (dA) is consistently zero due to symmetry, simplifying the dot product to (E.dA). As the electric field remains constant over the entire sphere, it can be factored out of the integral.
E . ∮dA = Qinclose/ε
Evaluate the Surface Integral:
- The surface integral (dA) equates to the surface area of the sphere, (4πr2).
E.4πr2 = Qinclose/ε
Solve for Electric Field:
- Solve for the electric field (E):
E = Q / 4πεr^2
Relate Electric Field to Coulomb's Law:
- Recognize that the electric field at a distance (r) due to a point charge (Q) aligns with Coulomb's Law:
[ E = kQ/r^2]
- Equate the expressions for (E) derived from Gauss's Law and Coulomb's Law:
[ Q / 4πεr2 = kQ/r^2 ]
- Cancel out (Q) and solve for the Coulomb constant (k):
[ k = 1/ 4πε ]
- This elucidates that Coulomb's Law emerges as a special case of Gauss's Law when considering a spherically symmetric charge distribution.
This meticulous derivation establishes the intricate relationship between Coulomb's Law and Gauss's Law, showcasing how the former is a distinct manifestation within the broader framework of the latter. The journey underscores the elegant interplay of these fundamental principles in the realm of electrostatics.
learn more :
electric field and application (in easy way )
electric force most important concepts with examples
charge density (linear, area, volume with most helpful important examples )
in-depth about charge with experiments
what is electric flux: know every things about it with iits examples