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- Electric flux through hemisphere - Physics Stack Exchange
- How can electric flux be negative? - Physics Stack Exchange
- Electric Flux through the Face of a Cube - Physics Forums
- electromagnetism - Normal Vector in Electric Flux and Gauss's …
- Electric flux calculation in case of a cylinder - Physics Forums
- Why does electric flux have 'cos θ' in its formula? - Physics Forums
- electrostatics - What is Electric flux? - Physics Stack Exchange
- Net electric flux through torus - Physics Forums
- Electric flux, vector or scalar? - Physics Forums
- Electric Flux through a hemisphere - Physics Forums
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In electromagnetism, electric flux is the total electric field that crosses a given surface. The electric flux through a closed surface is equal to the total charge contained within that surface.
The electric field E can exert a force on an electric charge at any point in space. The electric field is the gradient of the electric potential.
Overview
An electric charge, such as a single electron in space, has an electric field surrounding it. In pictorial form, this electric field is shown as "lines of flux" being radiated from a dot (the charge). These are called Gauss lines. Note that field lines are a graphic illustration of field strength and direction and have no physical meaning as isolated lines. The density of these lines corresponds to the electric field strength, which could also be called the electric flux density: the number of "lines" per unit area. Electric flux is directly proportional to the total number of electric field lines going through a surface. For simplicity in calculations it is often convenient to consider a surface perpendicular to the flux lines. If the electric field is uniform, the electric flux passing through a surface of vector area A is
Φ
E
=
E
⋅
A
=
E
A
cos
θ
,
{\displaystyle \Phi _{\text{E}}=\mathbf {E} \cdot \mathbf {A} =EA\cos \theta ,}
where E is the electric field (having the unit V/m), E is its magnitude, A is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to A.
For a non-uniform electric field, the electric flux dΦE through a small surface area dA is given by
d
Φ
E
=
E
⋅
d
A
{\displaystyle {\textrm {d}}\Phi _{\text{E}}=\mathbf {E} \cdot {\textrm {d}}\mathbf {A} }
(the electric field, E, multiplied by the component of area perpendicular to the field). The electric flux over a surface is therefore given by the surface integral:
Φ
E
=
∬
S
E
⋅
d
A
{\displaystyle \Phi _{\text{E}}=\iint _{S}\mathbf {E} \cdot {\textrm {d}}\mathbf {A} }
where E is the electric field and dA is an infinitesimal area on the surface with an outward facing surface normal defining its direction.
For a closed Gaussian surface, electric flux is given by:
where
E is the electric field,
dA is an infinitesimal area on the closed surface,
Q is the total electric charge inside the surface,
ε0 is the electric constant (a universal constant, also called the permittivity of free space) (ε0 ≈ 8.854187817×10−12 F/m)
This relation is known as Gauss's law for electric fields in its integral form and it is one of Maxwell's equations.
While the electric flux is not affected by charges that are not within the closed surface, the net electric field, E can be affected by charges that lie outside the closed surface. While Gauss's law holds for all situations, it is most useful for "by hand" calculations when high degrees of symmetry exist in the electric field. Examples include spherical and cylindrical symmetry.
The SI unit of electric flux is the volt-meter (V·m), or, equivalently, newton-meter squared per coulomb (N·m2·C−1). Thus, the unit of electric flux expressed in terms of SI base units is kg·m3·s−3·A−1. Its dimensional formula is L3MT−3I−1.
See also
Magnetic flux
Maxwell's equations
Electric field
Magnetic field
Electromagnetic field
Citations
References
Purcell, Edward; Morin, David (2013), Electricity and Magnetism (3rd ed.), Cambridge University Press, New York, ISBN 9781107014022
Browne, Michael (2010), Physics for Engineering and Science (2nd ed.), McGraw Hill/Schaum, New York, ISBN 0071613994
External links
Electric flux – HyperPhysics
Kata Kunci Pencarian:
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Electric flux - Physics
electric flux
Daftar Isi
Electric flux through hemisphere - Physics Stack Exchange
Aug 30, 2020 · The electric flux through the curved surface area of a hemisphere of radius R when it is placed in a uniform electric field is? Before this, I was taught the definition of flux as the number of field lines passing perpendicularly through an area. (If the lines aren't perpendicular, we use the component of field line that is)
How can electric flux be negative? - Physics Stack Exchange
Apr 24, 2020 · Typically in this case we define "outward" flux to be positive, in which case the total electric flux through the surface will be a positive number since the electric field lines point out of a positively charged particle. If we repeated the same experiment with a negatively charged particle we would get a negative answer.
Electric Flux through the Face of a Cube - Physics Forums
Jun 24, 2018 · The flux into the volume in this corner must then be equal to the flux out of the three sides away from the centre and can be easily computed to be the solid angle covered by the corner (##\pi/2##) divided by ##4\pi##.
electromagnetism - Normal Vector in Electric Flux and Gauss's …
Aug 9, 2021 · In the case where the surface "encloses a volume" $\Omega$, such as the surface of a sphere, which is the boundary of a ball, we always choose $\mathbf{n}$ to "point outwards" because then, the divergence theorem states \begin{align} \int_{\Omega}\text{div}(\mathbf{F})\,dV&=\int_{\partial \Omega}\mathbf{F}\cdot \mathbf{n}\,dA …
Electric flux calculation in case of a cylinder - Physics Forums
Sep 13, 2015 · an electric field is uniform,and in the positive x direction for positive x,and uniform with the same magnitude but in the negative x direction for negative x.it is given that vector E=200 I^ (N/C) for x>0 and vector E= -200 (N/C) for x<0 .A right circular cylinder of length 20 cm and radius 5 cm has it's center at the origin and it's axis ...
Why does electric flux have 'cos θ' in its formula? - Physics Forums
Mar 7, 2015 · To find the flux through a surface you need the component of the field perpendicular to the surface. Taking that component involves the cosine of the angle between the field and the normal to the surface. Read about it here: Electric Flux
electrostatics - What is Electric flux? - Physics Stack Exchange
Jan 28, 2025 · Electric or Magnetic flux usually come up when considering Faraday and Ampere laws in Maxwell's equations, and appear explicitly in their integral form. It also comes up when trying to calculate enclosed charge in a region using Gauss Law .
Net electric flux through torus - Physics Forums
Jan 29, 2007 · What is the net electric flux through the torus (i.e., doughnut shape) of the figure Homework Equations net flux= E*A I believe is needed The Attempt at a Solution I don't know how to do this problem at all. I feel like I don't have enough information to calculate the answer.
Electric flux, vector or scalar? - Physics Forums
Dec 11, 2012 · The electric flux density vector is used to calculate the electric flux passing through any and all arbitrarily oriented cross sectional areas dA in space. Of course, for a given electric flux density vector, the electric flux passing through a given surface area will depend on how the surface area is oriented in space.
Electric Flux through a hemisphere - Physics Forums
Aug 30, 2007 · Homework Statement "What is the flux through the hemispherical open surface of radius R? The uniform field has magnitude E. Hint: Don't use a messy integral!" \\mid \\vec{E} \\mid= E radius = R Homework Equations Electric Flux over a surface (in general) \\Phi = \\int \\vec{E} \\cdot \\,dA...