In physics and electromagnetism , Gauss's law, also known as Gauss's flux theorem, (or sometimes simply called Gauss's theorem) is a law relating to the distribution of electric charge to the resulting electric field. In its integral form, it states that the flux of the electric field out of an arbitrary closed surface is proportional to the electric charge enclosed by the surface, irrespective of how that charge is distributed. Even though the law alone is insufficient to determine the electric field across a surface enclosing any charge distribution, this may be possible in cases where symmetry mandates uniformity of the field. Where no such symmetry exists, Gauss's law can be used in its differential form, which states that the divergence of the electric field is proportional to the local density of charge.
The law was first formulated by Joseph-Louis Lagrange in 1773, followed by Carl Friedrich Gauss in 1813, both in the context of the attraction of ellipsoids. It is one of Maxwell's four equations, which forms the basis of classical electrodynamics. Gauss's law can be used to derive Coulomb's law, and vice versa.
I tried to solve the question using two different approaches to gain a better understanding of the subject. However, I reached two different results with each approach.
I believe I used Gauss's law to find the electric charge distribution and the electric field inside the cavity incorrectly...
To apply the Divergence Theorem (DT), at least as it is stated and proved in undergrad calculus, it is required for the vector field ##\vec{F}## to be defined both on the surface ∂V, so that we can evaluate the flux through this surface, and on the volume V enclosed by ∂V, so that we can...
My solution is this:
$$q = \varepsilon_0 \int E.dA$$
Based on gauss's law.
Taking the derivative of both sides with respect to $$A$$ we get:
$$\frac{dq}{dA} = \varepsilon_0 E$$
From chain rule:
$$\frac{dq}{dA} = \frac{\frac{dq}{dr}}{\frac{dA}{dr}}$$
On the other hand:
$$q = \int \rho dv = \int...
I have read Griffiths' Chapter 2 sections on Conductors. According to it, (if I understood it correctly) if the charge is put inside the cavity of a conductor, then the equal and opposite total charge will be induced surrounding the cavity. This charge and the total charge induced surrounding...
This is a problem from Yale OCW (Shankar). The solution he gives is as follows:
Sure, this makes sense. However...
Superimposed rho and negative rho with radius R/2 means there is no charge enclosed in the cavity... therefore
no charge -> no flux -> no electric field.
Hi, I would very much appreciate some guidance on the below.
Consider a one-dimensional world as depicted in the attached figure.
We have two (lets say positively charged particles enclosed by two conductor plates.
One plate is at ##x=0##, the other at ##x=L##. The particles are at ##x_1## and...
So the question is like I mentioned. I know how to use GL and ACL in place of CL and BSL. I also know that they make the calculations simpler.
The problem is whenever GL and ACL are used in place of CL or BSL, they are used in the following way:
Gauss Law:$$\int{ E...
This is a conceptual question. I think we can conclude that electric charges cannot be present if there is no electric field in that region. Is this an application of Gauss' Law? A net electric flux thru a surface indicates that there is a charge within that region. An electric field must be...
Okay so I am a little confused as to where I made a mistake. I couldn't figure out how to program Latex into this website but I attached a file with the work I did and an explanation of my thought process along the way.
I know the Gauss law for surface integral to calculate total charge by integrating the normal components of electric field around whole surface . but in above expression charge is calculated using line integration of normal components of electric field along line. i don't understand this...
I solved laplacian equation. and got the solution of V(r, phi) = a. +b.lnr + (summation) an r^n sin(n phi +alpha n ) + (summation) bn r ^-n sin( n phi +beta n)
When I try to derive Gauss's law with a straight line of charge with density ##\lambda## through a cylindrical surface of length L and radius R,
$$\vec E = \frac{\lambda*L}{4\pi\epsilon*r^2}$$
$$A = 2\pi*r*L$$
$$\vec E*A = \frac{\lambda *L^2}{2\epsilon*r} \neq \frac{q_{enc}}{\epsilon}$$
What am...
Draw a Gaussian pill box that starts from 0 (half way between the slab) and extends towards 2 cm.$$A \times \int_{0}^{0.02} \rho dz$$
I'm not sure if I should multiply the integral by A (area) or V (volume)
And if area would I multiply by 0.02^2?
I'm confused here. Thanks for your help.
First draw a gaussian shape outside of the sphere (a larger sphere) with radius R. The total charge from the (inner) sphere will be:
$$Q = \sigma A$$
$$A = 4\pi r^2$$
$$Q = \sigma 4\pi r^2$$
Use Gauss's Law to derive electric field magnitude
$$\oint_{}^{} E \cdot dA = \frac{q_e}{\epsilon_o}$$...
Picture :
My answer :
I guess net electric flux is 0.
so electric flux passing through surface 1 = -(electric flux passing through surface 2)
and electric flux passing through surface 1 is EA = E(pi)(r^2)
Is it correct? Thank you ...
Gauss' law: $$\iint_{\partial A}\vec E\cdot d\vec A=\frac{Q}{\epsilon_0}$$
Suppose we have a unevenly charged non-conducting spherical shell, in which a Gaussian surface is placed. In this case, is the electrical field on A 0, given that there is no charge inside A? I came up with this example...
confused on part A/B when I look up they did E= Q/2e(0.8)^2.
But why not use the 0.100mm because that is the area of the enclosed.
Same with B why did they use 100m and not 0.8m because 0.8 is smaller so it enclosed the charge
Electric Flux = E*A = 5*6(0.05)^2.
when i look up at other sources they use Electric flux = q/ (8.854*10^-12 [this is e]) equation
but I am confused on why the E*A equation don't work. The answer is 0.02Nm^2/C
What am I missing?
I also don't get the title of the section: "Charge distributions with enough symmetry for Gauss's Law".
I thought Gauss's Law was valid for any closed surface enclosing a charge. I don't understand what "enough symmetry" means in the title above. I get that with symmetry...
Hi guys i have some concept issues about flux. My book says flux is proportional to the number of lines passing through that area. so my question is: first i am thinking that a 2D circle which has a 2r diameter and it is enclosing a 2q charge so its flux should be 2q/Epsilon.then i am thinking a...
Gauss law relates the net flux phi of an electric field through a closed surface to the net charge q that is enclosed by that surface. It tells us that
Phi = q/permittivity
Can I say it like this : The gauss law states that the net flux of the surface depends upon the net charge enclosed by that...
My book claims that the diff. form of Gauss' law is
$$\nabla\cdot\mathbf E=4\pi\rho$$
Can someone tell me why it isn't ##\nabla\cdot\mathbf E=\rho/\epsilon_0##?
Hi.
I was reading about conductors in electrostatic equilibrium and how it makes sense that they have zero electric field inside the material even when an external charge is brought near. The charge density of the material just rearranges itself to cancel. Then I searched for hollow conductors...
Hello everybody
To calculate the flux for the electric field I need the gauss law. There are two formula one with the integration over some area and the other is Q/e0. When do I have to use which one?
I am trying to derive that
$$\nabla \times B=\mu_0 J$$
First the derivation starts with the electric field
$$dS=rsin\varphi d\theta r d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0} \iint\limits_S \frac{r}{|r|^3} \cdot dS $$...
I'm a bit confused on the derivation above. I understand what the goal of the derivation is, as it derives Gauss's Law using the solid angle, but i was wondering if someone could kind of fill in the steps the author skipped and explain the use of the solid angle.
F = qE
ma = (2*10^-6) * (λ / (2pi*r*ε0) )
ma = (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) => I am not certain what to put for r ( But I sub in 4 because dist is 4)
a = ( (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) )/ 0.1
a = 0.35950
v^2 = U^2 + 2 a s
v = 0
u^2 = -2 a s => Can't sqrt negative so...
Attached is problem 23.03 from Halliday and Resnick.
We have a sphere of uniform negative charge Q = -16e and radius R = 10cm. at the center of the sphere is a positively charged particle with charge q = +5e. We are supposed to use Gauss' law to find the magnitude of the electric field at...
Gauss' law dictates that charge will only appear on the outer surface of a conductor. But if there's charge in a conducting cavity, the inner surface of the conductor will accumulate induced charge. So what's outer or inner should be redefined?
In derivations of capacitance it is standard to consider two oppositely charged, infinitely thin sheets. If we construct a Gaussian cylinder across one sheet, we obtain ##E_{1} = \frac{\sigma}{2\epsilon_{0}}## for one sheet, and then we can superpose this field with that from the other at an...
This is my attempt, i am confused at some points
a. r = 0; The Electric field is 0
b. At r = a/2.00; I verified the answer and it is non zero, but my understanding is that the net charge should be on the surface of the conductor. Hence the charge q1=5*10^-15 C, should go to the surface of the...
I really don't understand the theory of the above kind of questions. But from the little theory i understand the Electric field is 0 inside the conductor and all the charge goes to the surface and distributes equally.
a. Since the E=0 inside the conductor the point charge distributes outside...
i have little experience with the differential form of Gauss's Law, and I've tried three times now to arrive at it for a point mass M (spherically symmetric classical gravitational field) but instead of getting an answer proportional to the mass density I keep getting zero. Is the divergence...
I have read multiple threads on Physics Forums, Stackexchange and Quora, as well as the explanation of Gauss Law, but still don't understand the most fundamental aspect of it: its applicability for any kind of surface. More precisely, I don't get how this follows from the fact that...
I've just been learning about Gauss' law which as far as I can tell states that the net electric flux through a surface equals the enclosed charge divided by the permittivity of free space, and is often expressed as the integral $$\int_S {\bf{E} \cdot d \bf{A}} = \frac{Q}{\epsilon_0}$$In some...
Hi,
I'm trying to prove Gauss's Law by using a cubical surface with a point charge located at its center, and I'm running up against some difficult integration. I've worked through the first integral of the surface integral, but I can't seem to figure out a proper integration technique. Here is...
So the first problem stated is to show that for a charge distribution between two spherical shells of radii r1<r2, the total charge inside is described by:
This is rather trivial using Gauss' law in integral form, so I regard this as completed.
I have used the gradient to find the electrical...
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so in the 2nd page,when the dielectric material is introduced the gauss's law becomes $$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { \epsilon _ { 0 } }$$.But my question is why the ##{ \epsilon _ { 0 } }## is in the equation.Shouldn't...
I tried to work out both a) and b), but I am not sure if I am correct. I drew a picture with a sphere around q first with radius r and then with radius 3r.
For a) ##E.A=\frac {q}{ε_°}## (when using Gauss' Law)
Since ##A=4πr^2##, I substituted this in the equation and solved for E giving me...
Which is better to use? The equation for the area or the circumference of a circle?
Schaum's Electromagnetics (4 ed) by Edminister
vs
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html
I know how Gauss law helps us to calculate the discontinuity at a point on the surface of a surface charge.
Similarly using Gauss law, is there a way to determine the continuity at other points of electric field due to a surface charge or the continuity at all points of electric field due to a...
Why does a Faraday Cage work? (or more generally, why does the inside of a conducting shell have no electric field if there are charges placed outside it?)
I understand that this is the result of polarisation, but why does the polarisation happen to exactly cancel out the field? Could there be...
Homework Statement
Given two things spherical shells radii r1 and r2 with r2 > r1.
The inner she'll is charged uniformly with a total charge Q1, while the outer shell with Q2.
A) use gauss law to computer the electric field everywhere
B) Use any method to calculate the potential everywhere...
Homework Statement
determine the electric flow through a square surface of side 2l due to a load + Q located at a perpendicular distance l from the center of the plane
I really don't know how to answer this question .i need help guys
Thanks
Homework EquationsThe Attempt at a Solution
I ended...
Homework Statement
A) use gauss's Law to determine the electric field at all values of radial distance (0<r<infinity) from the center of a non-uniformly charged cylinder that is very very long and lies along the x-axis. The cylinder carries excess charge per volume ρ=[a]r^2 (the [] are supposed...
Homework Statement
We have a uniformly charged, non-conducting sphere (charge per volume,ρ, and radius, R). Then a uniformly charged ruler (charge per length,λ, and length, d) is aligned radially almost touching the surface of the sphere. Determine the net force experienced by the ruler...
"Gauss's theorem can be established as follows. consider an attracting particle of mass m at the point P, and let a cone of small solid angle ω be generated by radii through P. This cone cuts the surface at the points Q1, Q2, ... taken in order from P; the parts of the surface cut off by the...
Homework Statement
You have a conducting sphere that is in equilibrium, it has a cavity in it with positive charge +q. If you bring another charge +q2 near the outer edge of the conductor does the total surface charge on the wall of the cavity, q(int) change? There is an image attached that...