Hello professor, From 5:32, you have asked why the E field is independent of h (distance), given the plane geometry is infinitely large. Reason 1: The plane is so large and given role (charge per unit volume) is uniform across, the amount of charge in the plane is so great that the Electric field changes with respect to small distance is negligible. Reason 2, we know field is inversely proportional to r squared and so for short distance is ok, but cannot generalise to large distances ( I think this is not a valid explanation, but decided to put it down anyway) Can you please explain if both of these are incorrect. Thank you
Why is the electric field due to infinite sheet of charge independent of the distance from it? (possible answer) The electric field at a point in space due to a charge distribution depends on the shape of the surface on which the charge is distributed (as one of the factor) Imagine that a person is trying to measure an electric field due to a U charged circular disc (or any other shape). initially the person is close to disc, so the person does 'see' it as a disc, so the electric field measured by him at that region will be that of a disc. But when the person is very very far away from the disc, he does not 'sees' it as a disc but just as a point so the electric field measured by him will be that of a point charge or, to put it in another way the electric field measured by him at any distance should depend upon the 'APPARENT SHAPE' of the surface viewed from that distance from it which is only possible for surface of finite size, for a surface like an infinite sheet of charge no matter how far or how close one is form an infinite sheet of charge it still appears as an infinite sheet of charge so the electric field is always same at all distance from it.
Sir the cylinder that you chose is making R redius on z axis. On the outer point of cylinder (on z axis at r distance away from surface). You must find a electric field at direction of z axis (due to charge inside the cylinder) and ds (area vector) is also on z axis. So they mac 0° angle. They have to make a dot product. So why you say that electric field on Z axis( vertex of cylinder or box) is zero. Please explain. In hope of explanation.
hello professor, we r applying the guass law here assuming the surface will only rediate electric field perpendicular to plane of slab right is our but is our assumtion right for long distances
What if you divide the slab horizontally in the middle in to two parts, one with positive charge and the other with equal, but negative charge? How will the electric field look, where will it be the strongest and weakest (0)? Specifically the centerline and the negative charge density slab confounds me. Spontaneously it feels like the E-field should be the strongest in the middle of the negatively charged slab (because the +field will try to push a positive charge q away, but the -field will try to pull it back, so the middle seems nice because then the +field has pushed as much as it could, while both ends of the -field will pull with equal force), but that doesn't seem like a symmetrical answer to a symmetrical problem.
This is a problem in SBT (an IIT JEE book) and I want to tell you, The ELECTRIC FEILD outside of that slab will be 0 and inside from the central line will be proportional to the displacement.
Right, however, if the distances at which you calculate the E-fields in or near the slab are very small compared to the size of the width (and if you stay away from the edges), your results can be very accurate to the point that your assumption of an infinitely wide slab is more than justified.
Gauss Law requires that you have a VOLUME. We chose the two flat parts at equal distance from the slab. Thus we know that the E-fields at the 2 flat parts are the same but in opposite direction. We now apply Gauss law and calculate how much charge there is in the volume and we find the E-field that we were looking for.
so we should choose a suitable volume to but the charge in then we determine the flux throw this volume and with it we can get the electric field easily ......................... and in cylindrical shape we have three areas the top and bottom and in between and we should get the electric field passing through them to get the flux Mmmmmmmmmmmmm i think i got it :D Thanks a lot you beauty ^^
sir one thing i couldn't understand that is the electric field calculated is only due to the charges included inside the cylinder or the whole body. If the later is true how is it possible?(sorry for my ignorance,i am in 12th grade)
Gauss' Law!! I agree it's not intuitive. It took Newton 40 years to prove that inside a spherical shell of uniform mass density there is NO gravitational field. The same is true for charges and E-fields. I suggest you google "divergence theorem".
Approximately at 6:40, where you say that you calculate the flux through the cylinder inside the slab and you put that equal to the charge inside divided by epsilon_zero. You then give us a non zero answer, so the charge inside can't be zero. But since it's a conductor the charge should be all at the surface no?
Hello professor,
From 5:32, you have asked why the E field is independent of h (distance), given the plane geometry is infinitely large.
Reason 1: The plane is so large and given role (charge per unit volume) is uniform across, the amount of charge in the plane is so great that the Electric field changes with respect to small distance is negligible.
Reason 2, we know field is inversely proportional to r squared and so for short distance is ok, but cannot generalise to large distances ( I think this is not a valid explanation, but decided to put it down anyway)
Can you please explain if both of these are incorrect. Thank you
Watching ur vedio today and u are looking very young Sir than today 💓😁💓
Thank u for making physics easy.😊
Now I can understand Physics easily
Only by hiiiiiiim
Why is the electric field due to infinite sheet of charge independent of the distance from it? (possible answer)
The electric field at a point in space due to a charge distribution depends on the shape of the surface on which the charge
is distributed (as one of the factor)
Imagine that a person is trying to measure an electric field due to a U charged circular disc (or any other shape). initially the person is close to disc, so the person does 'see' it as a disc, so the electric field measured by him at that region will be that of a
disc. But when the person is very very far away from the disc, he does not 'sees' it as a disc but just as a point so the electric field measured by him will be that of a point charge or, to put it in another way the electric field measured by him at any
distance should depend upon the 'APPARENT SHAPE' of the surface viewed from that distance from it which is only possible for
surface of finite size, for a surface like an infinite sheet of charge no matter how far or how close one is form an infinite sheet
of charge it still appears as an infinite sheet of charge so the electric field is always same at all distance from it.
watch my lecture on this topic (with demo) or use google.
This is very basic.
First clearify yourself that whether you are asking a question or answering it !
Hello Professor Lewin, is the answer to the electric field due to a slab of charge for x< d = px/e_o ?
+Dr. Science Sc.D correct
how did you find it because I am getting it as pd/e_o
Sir the cylinder that you chose is making R redius on z axis. On the outer point of cylinder (on z axis at r distance away from surface). You must find a electric field at direction of z axis (due to charge inside the cylinder) and ds (area vector) is also on z axis. So they mac 0° angle. They have to make a dot product. So why you say that electric field on Z axis( vertex of cylinder or box) is zero.
Please explain.
In hope of explanation.
I choosen z axis as out side from your paper.
hello professor, we r applying the guass law here assuming the surface will only rediate electric field perpendicular to plane of slab right is our but is our assumtion right for long distances
At the beginning you said it was a plane but inmediatly you say the density is in m^3 why is that? shouldnt be m^2?
how many minutes into the video?
0:43
the plane is infinitely large with a thickness 2d.
Ok yes I understand it now. Thanks for such a fast answer :)
What if you divide the slab horizontally in the middle in to two parts, one with positive charge and the other with equal, but negative charge? How will the electric field look, where will it be the strongest and weakest (0)? Specifically the centerline and the negative charge density slab confounds me. Spontaneously it feels like the E-field should be the strongest in the middle of the negatively charged slab (because the +field will try to push a positive charge q away, but the -field will try to pull it back, so the middle seems nice because then the +field has pushed as much as it could, while both ends of the -field will pull with equal force), but that doesn't seem like a symmetrical answer to a symmetrical problem.
work it out - my lectures will help you.
This is a problem in SBT (an IIT JEE book)
and I want to tell you, The ELECTRIC FEILD outside of that slab will be 0 and inside from the central line will be proportional to the displacement.
I'm studying law. I have no idea why I am here.
ahahaha, depends on which law u studying
Sir i am From India
I'm confused about how the gaussian surface is oriented with respect to the slab...
watch the video again
The charged object is not a plane but a slab whose dimensions are (2d) x ∞ x ∞. The cylinder's axis is paralled to (2d) dimension. I hope I help.
Is the slab infinitely wide? If no, we do not have this solution, right?
Right, however, if the distances at which you calculate the E-fields in or near the slab are very small compared to the size of the width (and if you stay away from the edges), your results can be very accurate to the point that your assumption of an infinitely wide slab is more than justified.
Is the solution correct sir?
U doubt ?? 🧐
i don't understand why did we take a consideration of the other area on the other side of the slab
we only want to get the electric field on one side
Gauss Law requires that you have a VOLUME. We chose the two flat parts at equal distance from the slab. Thus we know that the E-fields at the 2 flat parts are the same but in opposite direction. We now apply Gauss law and calculate how much charge there is in the volume and we find the E-field that we were looking for.
so we should choose a suitable volume to but the charge in then we determine the flux throw this volume and with it we can get the electric field easily
.........................
and in cylindrical shape we have three areas the top and bottom and in between
and we should get the electric field passing through them to get the flux
Mmmmmmmmmmmmm
i think i got it :D
Thanks a lot you beauty ^^
Symmetric argument
sir one thing i couldn't understand that is the electric field calculated is only due to the charges included inside the cylinder or the whole body. If the later is true how is it possible?(sorry for my ignorance,i am in 12th grade)
Gauss' Law!! I agree it's not intuitive. It took Newton 40 years to prove that inside a spherical shell of uniform mass density there is NO gravitational field. The same is true for charges and E-fields. I suggest you google "divergence theorem".
thnx sir.
Why isn't the charge inside the slab in this case 0? Shouldn't it be only at the surface of the slab too?
how many minutes into the video?
Approximately at 6:40, where you say that you calculate the flux through the cylinder inside the slab and you put that equal to the charge inside divided by epsilon_zero. You then give us a non zero answer, so the charge inside can't be zero. But since it's a conductor the charge should be all at the surface no?
what is the name of the video (be as precise as possible), how many minutes into the lecture?
The name of the video is 8.02x - Module 01.04 - A Slab of Charge. It's from 6:05 to 7:00
The link to it is ua-cam.com/video/nW7LDCKWPqg/v-deo.html
it's not a conductor. charge is uniformly distributed throughout the slab.
Nice explanation but
Poor Pages 😂😂
is the solution of the problem asked at 7minutes into the video ,E=rho*x/epsilon
I got the same result, I think it is right
Can somebody tell me what is the difference between "charge density" and "charge distribution" ?
Thank you,know their exact meanings, I just wonder how to calculate the distribution. Of course, I will check my DICTIONARY
Density is C/m^3. Distribution is the way the charges are distributed on a surface or inside a insolator
sir, you skip too many steps and formulas