Mr. Biezen, I honestly watch your videos as soon as I get home from class. I leave class in a daze and you clear up all the pettifog and confusion. THANK YOU
Professor Van Biezen, thank you so much for these videos. Your lectures have helped me so much in keeping up with my Physics 2 course at the University of Pennsylvania. I can honestly say that I have learned more from you than my professor! Keep doing what you're doing!
"Liking" before watching every other video you have. I have been watching your Gauss's law videos so far and I can't thank you enough for simplifying this topic. I'm gonna ace my midterm!!
This is absolutely amazing. Every small thing I did not understand about this is explained. They usually skip over so many details that makes this so hard. Thank you!
Que : A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.
I am from Poland and I have a huge problem with English, but I've found you and what is better I understand you. So finally I feel prepared for my classes in the theory of electromagnetic field. Thank you so much!
Dear sir Biezen, May I ask if the formula used for when the Gaussian surface is smaller than the volume charge ( r < R) can also be used not just for spheres but also for cylinders and other volumetric shapes? Thank you so much in advanced.
That is correct. The electric field only depend on the amount of charge inside the Gaussian surface and it can be applied for cylinders, slabs, and spheres.
Hello, Mr. Michel Biezen, Greetings from a teeny tiny city (Imphal) in India! I usually look up your videos (plus others like Prof. Shankar's- Yale Univ & Prof Lewin-MIT) whenever I have troubles in electrostatics. I have a small question here (and this was actually asked by a student of mine) which is interesting. When we use the Guass's Law for the region inside the sphere and knowing the sphere is a non-conductor so has got some permittivity other than that of free-space value, why do not we use this material permittivity instead of the free-space permittivity? Awaiting for your reply. Thanks in advance!
Kamaljit,Tell your student that is a very good question. If the electric field inside the uniformly charged sphere, affects the dielectric material by polarizing the material and thus setting up an electric field in the reverse direction, the net electric field will be determined by not using the permittivity of free space, but the effective permittivity. In most "text book" examples this is not considered (and may not have to be if no polarization exists). Thus we still use the permittivity of free space unless the problem specifically directs us not to. Polarization is not likely as there are charges distributed throughout the material and would cause polarization in all directions effectively canceling. Then we also have to consider the boundary condition at the edge of the sphere as there cannot be a discontinuity and thus once outside the sphere the electric field outside must match the electric field inside the sphere at the boundary. Michel
Wouldn't the charge outside of the Gaussian surface, but inside the sphere affect the electric field. My intuition tells me that if a small point charge was put at the Gaussian surface it would be pushed inward by charge outside of Gaussian surface, but pushed outward by the charge within the Gaussian surface.
Zain, If the charge is distributed uniformly, then if you draw a Gaussian surface inside the sphere, the amount of charge contained inside that Gaussian surface can be found by calculating the ratio of the volume contained in the Gaussian surface to the volume of the whole sphere. For example, if the volume contained in the Gaussian surface is 1/3 the volume of the sphere it will contain 1/3 the charge of the whole sphere.
Michel van Biezen thankss alott I really appreciate that... I would humbly request u if u can make some videos about electric flux and electric field density will be a great help for my studies....
Zain -ul-aabdin I don't have much on electric flux, but you may want to look in this playlist for a better understanding E&M radiation: PHYSICS 50 ELECTROMAGNETIC RADIATION and this playlist on the electric field. PHYSICS 36 THE ELECTRIC FIELD
Alejandro, When r > R, the charge inside the Gaussian surface is the whole charge on the sphere (not a fraction of the charge like in part a of the problem.)
Hello Professor thought the electric field inside the sphere would be zero because the charge would move because of the force from one another in till it reaches the surface killing the electric field inside ? or is that only for conductors. Thank you for all your videos
Could you clarify so the first example is where the charge is inside the gaussian surface but outside the charged center. However, the next example is that they are outside both the charged center and the gausian surface
@@MichelvanBiezen yes sir i know r is variable and change from case to case i mean i study by changing r so my way it is correct or not .. and Thank very much
I'm a ;ittle confused. Is the sphere a solid like a billiard ball or hollow like a tennis ball? Either way, if the charge were a negative charge, would our direction of the field be in the other direction?
hi, I have a question about potential on the surface of a nonconducting sphere with uniform charge when the potential at its centre equals zero. Radius R = 0.022m Charge Q = +3.9fC I think Potential V at surface is V= (1/4*pi*epsilon0)(Q/R)=[1/4*pi*epsilon0][3.9x10^-15/0.022] = 1.59mV but the correct answer seems to be 0.80mV and I don't know what I'm doing wrong
I thought there is no E-field at all inside of a spherical shell or solid sphere. that the charge collects at the surface and the E-field exists only at the surface and away from the surface but never underneath the sruface
How would this change if the spherical object had a hollow center with no charge and the Gaussian object's radius was less than the radius of the spherical object, but greater then the radius of the hollow center. Would you use the same method.
Hi sir. I would like to ask why we didn't do the ratio thing for the second part of the problem, i mean when r>R. In the first part, we did the volume ratio but the second part we didn't. Can you make it little bit more clear? Thank you.
In the first part we only want to include the charge within the Gausian surface (which is just a fraction of the total). For the second part, since the Gaussian surface is outside the charges sphere you need to consider the total charge.
dear professor please explain the below problem. A point charge +10 micro coloumbs placed at a distance of 5 cm from the center of a conducting grounded sphere of radius 2 cm. what is the total induced charge on conducting sphere ?
hello Professor, unfortunately i could not catch the purpose in doing the ratio between the volume of gauss surface and the total surface. And ultimately what this ratio gives us ? thanks before .
You need to figure out the portion of the total charge that is inside the Gaussian surface. Not that Gauss's law requires you to consider only the charge inside the Gaussian surface.
Hey, great Video, but I have a question: If you have the E-Field of Something, how are you able to calculate the potential? I know, that E=-grad(phi), but how do i go the other way?
What would you do for enclosed charge if it depends on the radius. for example, you have a volumetric charge density p = k*r^2 where k is just a constant
Sir if we a charge q placed at the centre of a thin metallic spherical shell and a charge q1 placed at a distance r outside that sphere then what will be the force on that particle q which is inside the spherical shell ??
Hello!I have a midterm Thursday. I am really confused about this question. In this video,is it a insulating sphere? Is the electric field inside a conducting sphere will be zero always? Please reply me as soon as possible thankyou
Michel van Biezen Thank you for responding me. But I want to make one more thing. If it is given a conducting sphere with radius 3cm, and the question is what is electric field inside conducting sphere with radius 2cm? The answer is zero because it is an insulating sphere,right?
There is not enough information given to make a decision here. In general if you are inside a conducting sphere, there will be no electric field inside.
Hime Yoru I know I'm late with this response. The difference is this video deals with uniform volume while the other video (gauss law 3) deals with a conductor.
Sir we know that in case of conductors charges reside on the surface. So electric Field inside is zero... Example are metal.. Could plz gimme example of material having uniform charge density.. Could it be metal or any other material.
Any semiconductor material can be made to lack electrons giving it a positive charge density, or giving it extra electrons, giving it a negative charge density. Also insulating material can have extra charges or can be polarized causing an electric field to exist.
+LiftedArts This is where you want to separate intuition from the laws of physics. Some very smart people figured out that the electric field at any point (inside a Gaussian surface) only depends on the charge inside the Gaussian surface.
Students will ask question related to their homework via this means. We sometimes answer them if time permits. (There are lots of requests from around the world.;
0:20 very subtle but important detail: the sphere's NOT a conductor, hence there's charge in its inside
That is correct. With a conductor, all of the extra charge would reside on the outside surface.
I suggest to put this comment on the top.
so if it was a conductor the electric field on the inside (r
@@vekal3679 Yes because Q inside the enclosed area, which is the sphere with radius r would be zero.
@@rebah2631 alright, thank you very much :)
Mr. Biezen, I honestly watch your videos as soon as I get home from class. I leave class in a daze and you clear up all the pettifog and confusion. THANK YOU
I don’t understand why I pay for college when I learn more and better from these videos than from my professor
i know right..
same here. college is overrated
Only to get a paper in the end that certificates you went through training
to get connection between human, even theres only a little chance in it
would you have watched this video if you didn't study at college?
Professor Van Biezen, thank you so much for these videos. Your lectures have helped me so much in keeping up with my Physics 2 course at the University of Pennsylvania. I can honestly say that I have learned more from you than my professor! Keep doing what you're doing!
"Liking" before watching every other video you have. I have been watching your Gauss's law videos so far and I can't thank you enough for simplifying this topic. I'm gonna ace my midterm!!
I had problem in understanding this through my text book of H.R.K but really you simplified it. Hats off sir !!
This is absolutely amazing. Every small thing I did not understand about this is explained. They usually skip over so many details that makes this so hard. Thank you!
Glad it was helpful!
Thank you for your videos. They cover a breadth of content that most professors don’t!
You're very welcome!
This man makes everything way easier. Love it!
Glad to help
Que :
A hollow charged conductor has a tiny hole cut into its surface.
Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the
unit vector in the outward normal direction, and σ is the surface
charge density near the hole.
Thank you Mr. Biezen for making us more than prepared. All of your videos are amazing and they make every confusion of class clear....Thanks!!!!!!
I am from Poland and I have a huge problem with English, but I've found you and what is better I understand you. So finally I feel prepared for my
classes in the theory of electromagnetic field. Thank you so much!
Welcome to the channel. Great to hear we were able to help.
Gr8 video,tbh this 6 mins taught me a lot more than what I learn at the entrance coaching center,thx btw 😃
Dear sir Biezen,
May I ask if the formula used for when the Gaussian surface is smaller than the volume charge ( r < R) can also be used not just for spheres but also for cylinders and other volumetric shapes?
Thank you so much in advanced.
That is correct. The electric field only depend on the amount of charge inside the Gaussian surface and it can be applied for cylinders, slabs, and spheres.
Thank you so much! You’re videos are really helping me understand.
Glad they are helping.
we just need Teachers like you in our University
Thank you. We appreciate your comment. (What university do you attend?) 🙂
Really great video man. Watched 2 times to understand fully but really good.
My professor tried to explain this in two hours... and he failed, but now I understand the whole thing in SIX MINUTES...
Answer:
Part A - E = 133,333 N/C
Part B - E = 180 N/C
These videos are fantastic, really helpful with studying.
Glad you like them! 🙂
i am confused about 5:30 what do you mean charge outside is the same as charge inside
The charge inside the Gaussian surface equals the charge inside the sphere (when r > R)
when you're going over "r>R", shouldn't the appropriate subscript for "Q" be "Q_out" instead of "Q_in"?
No. With Gauss's law it is always Qin (Charge inside the Gaussian surface).
amazing lecture, Michael.
Hello, Mr. Michel Biezen,
Greetings from a teeny tiny city (Imphal) in India!
I usually look up your videos (plus others like Prof. Shankar's- Yale Univ & Prof Lewin-MIT) whenever I have troubles in electrostatics.
I have a small question here (and this was actually asked by a student of mine) which is interesting.
When we use the Guass's Law for the region inside the sphere and knowing the sphere is a non-conductor so has got some permittivity other than that of free-space value, why do not we use this material permittivity instead of the free-space permittivity?
Awaiting for your reply.
Thanks in advance!
Kamaljit,Tell your student that is a very good question. If the electric field inside the uniformly charged sphere, affects the dielectric material by polarizing the material and thus setting up an electric field in the reverse direction, the net electric field will be determined by not using the permittivity of free space, but the effective permittivity. In most "text book" examples this is not considered (and may not have to be if no polarization exists). Thus we still use the permittivity of free space unless the problem specifically directs us not to. Polarization is not likely as there are charges distributed throughout the material and would cause polarization in all directions effectively canceling. Then we also have to consider the boundary condition at the edge of the sphere as there cannot be a discontinuity and thus once outside the sphere the electric field outside must match the electric field inside the sphere at the boundary. Michel
why do i go to university if i can just do all your videos and you explain it 100x better?
Wouldn't the charge outside of the Gaussian surface, but inside the sphere affect the electric field. My intuition tells me that if a small point charge was put at the Gaussian surface it would be pushed inward by charge outside of Gaussian surface, but pushed outward by the charge within the Gaussian surface.
No, if the charge is outside the Gaussian surface the charge does not affect the electric field at the Gaussian surface.
Thank you so much finally everything is clear
Sir, thanks a lot for clear explanation.
how do we calculate the potential at the sphere center? What equation do we use?
V at the center = V at the surface + integral of E dr from the edge to the center.
@@MichelvanBiezen thank you!
thank you!!!! this video saved me and answered soooo many questions!!!!
i don not understand the volume ratio part. plz clarify me.. Thankss for such´great videos by the way..
Zain,
If the charge is distributed uniformly, then if you draw a Gaussian surface inside the sphere, the amount of charge contained inside that Gaussian surface can be found by calculating the ratio of the volume contained in the Gaussian surface to the volume of the whole sphere.
For example, if the volume contained in the Gaussian surface is 1/3 the volume of the sphere it will contain 1/3 the charge of the whole sphere.
Michel van Biezen thankss alott I really appreciate that... I would humbly request u if u can make some videos about electric flux and electric field density will be a great help for my studies....
Zain -ul-aabdin
I don't have much on electric flux, but you may want to look in this playlist for a better understanding E&M radiation:
PHYSICS 50 ELECTROMAGNETIC RADIATION
and this playlist on the electric field.
PHYSICS 36 THE ELECTRIC FIELD
Michel van Biezen Million times thanks Sir....;))
what values did you get?
rR = 1.8 x 10^4/r^2 N/C [rhat radially ourward]
anyone?
r
Sir u r best of all.........
Very well explained sir. Thanx.
Great videos with even more great explanation. I have a doubt in the second case (r>R). Why are we leaving Q alone and not Q * (4*pi*R^2) ???
Alejandro,
When r > R, the charge inside the Gaussian surface is the whole charge on the sphere (not a fraction of the charge like in part a of the problem.)
Ok I got that! But why is it different to this example? And thank you so much for your response. All sunday studying thanks to your videos. :)
That's the example!
dont we have to use the ratios of volumes if its outside as well just like we did with the G.S inside?
No, not when placing the Gaussian Surface outside the charged sphere. You just consider the total charge on the sphere.
Hello Professor thought the electric field inside the sphere would be zero because the charge would move because of the force from one another in till it reaches the surface killing the electric field inside ? or is that only for conductors. Thank you for all your videos
Yes that is how it is for conductors.
thank you
How can the electric field E(r) be constant when the charges inside the Gaussian surface have different distances from the Gaussian surface, Michel?
It can be shown that a spherical distribution of charge acts like all the charge is located at the center of the sphere.
Michel van Biezen ahh, thank you for responding, Michel :D
Could you clarify so the first example is where the charge is inside the gaussian surface but outside the charged center. However, the next example is that they are outside both the charged center and the gausian surface
What is your specific question?
Great videos.Thank you so much!!
Sir pls. U have r
Little "r" is used as the variable and represents any distance from the center to the edge. R is a constant and represents the radius of the sphere.
@@MichelvanBiezen yes sir i know r is variable and change from case to case i mean i study by changing r so my way it is correct or not .. and Thank very much
The method shown in the video is correct. If you get the same answer then your method is correct as well.
I'm a ;ittle confused. Is the sphere a solid like a billiard ball or hollow like a tennis ball? Either way, if the charge were a negative charge, would our direction of the field be in the other direction?
hi, I have a question about potential on the surface of a nonconducting sphere with uniform charge when the potential at its centre equals zero.
Radius R = 0.022m
Charge Q = +3.9fC
I think Potential V at surface is V= (1/4*pi*epsilon0)(Q/R)=[1/4*pi*epsilon0][3.9x10^-15/0.022] = 1.59mV
but the correct answer seems to be 0.80mV and I don't know what I'm doing wrong
The potential at the center of a sphere with uniform charge density is not zero.
@@MichelvanBiezen so is the question faulty or something? Is it not possible to calculate the potential on the surface of the sphere?
Yes, just place the Gaussian surface right around the surface of the object.
Thank you so much! I finally get it
thank you so much for those videos!
I thought there is no E-field at all inside of a spherical shell or solid sphere. that the charge collects at the surface and the E-field exists only at the surface and away from the surface but never underneath the sruface
Why the field inside the sphere isnt zero if the distribution of charge is symmetrical?
Note that the charge is distributed all throughout the sphere, not just on the surface only.
Sry i'm new in physics, why the vector E.dA is equal to E.dA?
and if you are close to the sphere what happens then? it starts to look like a flat surface instead of a point?
Regardless of the distance, the charge on the sphere will act as a point charge at the center of the sphere for positions outside the sphere.
How would this change if the spherical object had a hollow center with no charge and the Gaussian object's radius was less than the radius of the spherical object, but greater then the radius of the hollow center. Would you use the same method.
Donal Moloney yes but field at that point would be due to the charges present outside the Gaussian surface as well
Excuse me so what is d(density)? If I have only one R, Q, E and g then how can I find the "d"?
I believe you are referring to the charge density which is the amount of charge per unit volume. d = Q/V = total charge / total volume
Hi sir. I would like to ask why we didn't do the ratio thing for the second part of the problem, i mean when r>R. In the first part, we did the volume ratio but the second part we didn't. Can you make it little bit more clear? Thank you.
In the first part we only want to include the charge within the Gausian surface (which is just a fraction of the total). For the second part, since the Gaussian surface is outside the charges sphere you need to consider the total charge.
k, but how do i find electric energy stored, possibly inside?
The energy stored will either be as electric fields or magnetic fields. It this case with stationary charges, it will be electric field
@@MichelvanBiezen do i have to use the U=εE²τ/2? quite unusual not using it for a condensator
dear professor please explain the below problem.
A point charge +10 micro coloumbs placed at a distance of 5 cm from the center of a conducting grounded sphere of radius 2 cm. what is the total induced charge on conducting sphere ?
Good lecture, but like charges repel each other. That's why high voltage power lines should be hollow! Kevlar inner core! ;) "skin effect"
hello Professor, unfortunately i could not catch the purpose in doing the ratio between the volume of gauss surface and the total surface. And ultimately what this ratio gives us ? thanks before .
You need to figure out the portion of the total charge that is inside the Gaussian surface. Not that Gauss's law requires you to consider only the charge inside the Gaussian surface.
Hey, great Video, but I have a question: If you have the E-Field of Something, how are you able to calculate the potential? I know, that E=-grad(phi), but how do i go the other way?
Take a look at this playlist: PHYSICS 38 ELECTRICAL POTENTIAL
Sir, for the second part, would I still use the same charge ratio that was originally derived?
For the second part, Q inside is the total charge in the sphere since the whole sphere is no inside the Gaussian surface.
Got it. Thanks, sir. Greatly appreciated.
What is potential for the first sample?
hello Sir, unfortunately i couldn't understand your aim in finding the ratio(in part where r
That was done to find the amount of charge in the spherical region with radius r. (since the charge is uniformly distributed throughout the volume).
Michel van Biezen thank you sir )
so this is an insulator ?
+yassin alzaiem
Yes, with a conductor all of the charges would be at the surface.
What would you do for enclosed charge if it depends on the radius. for example, you have a volumetric charge density
p = k*r^2 where k is just a constant
We have a few videos that show you how to deal with charge densities that vary with radius, etc.
Sir if we a charge q placed at the centre of a thin metallic spherical shell and a charge q1 placed at a distance r outside that sphere then what will be the force on that particle q which is inside the spherical shell ??
zero force because K OF metallic conductor is infinity hence force= (q.q1)/(4pi epsnot .K(INFINITY) R.R)= q.q1/infinity =0.
so you only do V(gaussian surface)/V(sphere) when the radius of the gaussian surface is less than the object radius?
Yes, and if the charge distribution is uniform within the sphere.
Michel van Biezen Much appreciated, thank you!
Hello!I have a midterm Thursday. I am really confused about this question. In this video,is it a insulating sphere? Is the electric field inside a conducting sphere will be zero always? Please reply me as soon as possible thankyou
The answers are yes and yes. (If the sphere was a conductor, the excess charges would reside on the surface only)
Michel van Biezen Thank you for responding me. But I want to make one more thing. If it is given a conducting sphere with radius 3cm, and the question is what is electric field inside conducting sphere with radius 2cm? The answer is zero because it is an insulating sphere,right?
Because it is a conducting sphere. My mistake
Michel van Biezen this video belongs to insulating sphere or conducting sphere
There is not enough information given to make a decision here. In general if you are inside a conducting sphere, there will be no electric field inside.
So, if this sphere is a conductor not insulator, and r is less than R , then E=0?
Then all the charge would reside on the surface and you would be correct.
@@MichelvanBiezen Thank you for your response and videos.
Thank you so much!!!
You're welcome! 🙂
whats the difference between this and the other video (gauss law 3) they both seem to look for the electric field outside a sphere
Susan,
I updated the set at a later time and probably duplicated some.
Hime Yoru
Yes they are very similar.
Hime Yoru I know I'm late with this response. The difference is this video deals with uniform volume while the other video (gauss law 3) deals with a conductor.
Sir we know that in case of conductors charges reside on the surface. So electric Field inside is zero...
Example are metal..
Could plz gimme example of material having uniform charge density..
Could it be metal or any other material.
Any semiconductor material can be made to lack electrons giving it a positive charge density, or giving it extra electrons, giving it a negative charge density. Also insulating material can have extra charges or can be polarized causing an electric field to exist.
@@MichelvanBiezen so when we talk about uniformly charged sphere it's not metallic.
Since in case of metal charges reside on the surface..
that is correct
But why don't you have to find the ratios for when the Gaussian radius is larger than the object?
Because then all of the charge is within the Gaussian surface.
Nice sir
great help, thank uuuuu
inside a sphere there is no charge so the electiric field inside the sphere wouldnt be equal to zero?
If there is no charge inside the sphere, the electric field inside the sphere will be zero.
BUT the teacher told us electrical field=0 inside the conductor because the vectoe cancel each ather
That is correct for a conductor. This is therefore not a conductor.
For the 1st one: Wouldn't all charges cancel due to symmetry?
+LiftedArts
This is where you want to separate intuition from the laws of physics. Some very smart people figured out that the electric field at any point (inside a Gaussian surface) only depends on the charge inside the Gaussian surface.
Oh ok. Thanks Michael Michel van Biezen
good sir g
good thumbnail choices
who is disliking the video :(
neighbors who don't like us, ex girlfriends LOL or kkkkkkk from brazil
THANK YOU SO MUCHHH
Instruction unclear ended up getting a free amperemeter in my house now ,how?
Owsome!!!
Glad you liked it.
sir, i'd like to know the difference between part 3 and part 6... can you please explain
In this video, the charge in the sphere is distributed throughout the sphere (not just at the surface only).
Michel van Biezen thanks sir. youre the best
Thank you!
why do we use r vs. R?
Sana,
No particular reason other than clarity. To indicate that there are two different radii.
Got it!
Can you please go over the properties of insulators and conductors and how they effect charge?
Sana Sarfaraz
There are some videos in playlist PHYSICS 39 which goes over dielectrics in capacitors.
why 4/3 pi r^2 and not just 4 pi r^3?
Lili,
Did you mean to ask why 4 pi r^2 and not 4/3 pi r^3?
If so, one is the surface area of a sphere while the other is the volume of the sphere.
Michel van Biezen Got it, thank you for clearing that up :)
I thought Gauss's Law was always in terms of surface area not volume?
John,
You are correct.
(But you need the volume integral to calculate the charge inside the Gaussian surface.)
thanks
If we were given the charge density could we have solved it using integrals?
Yes, we have some videos showing how to do that.
You learn these things in college??,these are my 12th standard questions 🤧
Where do you live?
@@MichelvanBiezen in india sir,we don't actually have it in our books but teachers teach this because it come in exams
Yes the high school curriculum is advanced in Indian high schools.
@@MichelvanBiezen oh,thanks again for viedo sir
Hello,professor do you have email.I have 3 problems. I want you to help me. Please
Students will ask question related to their homework via this means. We sometimes answer them if time permits. (There are lots of requests from around the world.;
👍
fank yu
Thank you so much.