0:20 very subtle but important detail: the sphere's NOT a conductor, hence there's charge in its inside

@MichelvanBiezen

5 жыл бұрын

That is correct. With a conductor, all of the extra charge would reside on the outside surface.

@jonathansum9084

4 жыл бұрын

I suggest to put this comment on the top.

@vekal3679

4 жыл бұрын

so if it was a conductor the electric field on the inside (r

@rebah2631

4 жыл бұрын

@@vekal3679 Yes because Q inside the enclosed area, which is the sphere with radius r would be zero.

@vekal3679

4 жыл бұрын

@@rebah2631 alright, thank you very much :)

I don’t understand why I pay for college when I learn more and better from these videos than from my professor

@lontongtepungroti2777

6 жыл бұрын

i know right..

@dannggg

6 жыл бұрын

same here. college is overrated

@thiagovilla970

5 жыл бұрын

Only to get a paper in the end that certificates you went through training

@gilgamesh1245

4 жыл бұрын

to get connection between human, even theres only a little chance in it

@whateveryh2119

4 жыл бұрын

would you have watched this video if you didn't study at college?

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!!

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 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!

@MichelvanBiezen

2 жыл бұрын

Glad it was helpful!

Gr8 video,tbh this 6 mins taught me a lot more than what I learn at the entrance coaching center,thx btw 😃

Thank you for your videos. They cover a breadth of content that most professors don’t!

@MichelvanBiezen

3 жыл бұрын

You're very welcome!

This man makes everything way easier. Love it!

@MichelvanBiezen

3 жыл бұрын

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.

amazing lecture, Michael.

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!

@MichelvanBiezen

3 жыл бұрын

Welcome to the channel. Great to hear we were able to help.

Answer: Part A - E = 133,333 N/C Part B - E = 180 N/C

thank you!!!! this video saved me and answered soooo many questions!!!!

we just need Teachers like you in our University

@MichelvanBiezen

2 жыл бұрын

Thank you. We appreciate your comment. (What university do you attend?) 🙂

Great videos.Thank you so much!!

thank you so much for those videos!

These videos are fantastic, really helpful with studying.

@MichelvanBiezen

2 жыл бұрын

Glad you like them! 🙂

Thank you so much finally everything is clear

My professor tried to explain this in two hours... and he failed, but now I understand the whole thing in SIX MINUTES...

Thank you so much! I finally get it

Sir, thanks a lot for clear explanation.

Very well explained sir. Thanx.

Sir u r best of all.........

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 ?

great help, thank uuuuu

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!

@MichelvanBiezen

6 жыл бұрын

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

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

Thank you!

THANK YOU SO MUCHHH

Thank you so much.

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.

@vikramram7011

7 жыл бұрын

Donal Moloney yes but field at that point would be due to the charges present outside the Gaussian surface as well

Thank you so much!!!

@MichelvanBiezen

11 ай бұрын

You're welcome! 🙂

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 .

@MichelvanBiezen

7 жыл бұрын

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.

what values did you get? rR = 1.8 x 10^4/r^2 N/C [rhat radially ourward] anyone?

@Funnyketvideos

6 жыл бұрын

r

Sry i'm new in physics, why the vector E.dA is equal to E.dA?

when you're going over "r>R", shouldn't the appropriate subscript for "Q" be "Q_out" instead of "Q_in"?

@MichelvanBiezen

Жыл бұрын

No. With Gauss's law it is always Qin (Charge inside the Gaussian surface).

What is potential for the first sample?

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?

@MichelvanBiezen

6 жыл бұрын

Take a look at this playlist: PHYSICS 38 ELECTRICAL POTENTIAL

why do i go to university if i can just do all your videos and you explain it 100x better?

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

@MichelvanBiezen

5 жыл бұрын

We have a few videos that show you how to deal with charge densities that vary with radius, etc.

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

7 жыл бұрын

What is your specific question?

Dear sir Biezen, May I ask if the formula used for when the Gaussian surface is smaller than the volume charge ( r Thank you so much in advanced.

@MichelvanBiezen

5 ай бұрын

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.

@jorusenpai

5 ай бұрын

Thank you so much! You’re videos are really helping me understand.

@MichelvanBiezen

5 ай бұрын

Glad they are helping.

How can the electric field E(r) be constant when the charges inside the Gaussian surface have different distances from the Gaussian surface, Michel?

@MichelvanBiezen

7 жыл бұрын

It can be shown that a spherical distribution of charge acts like all the charge is located at the center of the sphere.

@HakaTech

7 жыл бұрын

Michel van Biezen ahh, thank you for responding, Michel :D

dont we have to use the ratios of volumes if its outside as well just like we did with the G.S inside?

@MichelvanBiezen

Жыл бұрын

No, not when placing the Gaussian Surface outside the charged sphere. You just consider the total charge on the sphere.

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.

@MichelvanBiezen

4 жыл бұрын

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.

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.

@MichelvanBiezen

4 жыл бұрын

No, if the charge is outside the Gaussian surface the charge does not affect the electric field at the Gaussian surface.

thanks

Excuse me so what is d(density)? If I have only one R, Q, E and g then how can I find the "d"?

@MichelvanBiezen

10 ай бұрын

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

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 ??

@umashankaryadav8576

7 жыл бұрын

zero force because K OF metallic conductor is infinity hence force= (q.q1)/(4pi epsnot .K(INFINITY) R.R)= q.q1/infinity =0.

Owsome!!!

@MichelvanBiezen

2 жыл бұрын

Glad you liked it.

Nice sir

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

@MichelvanBiezen

7 жыл бұрын

Yes that is how it is for conductors.

@zamirbarbosa8223

7 жыл бұрын

thank you

and if you are close to the sphere what happens then? it starts to look like a flat surface instead of a point?

@MichelvanBiezen

5 жыл бұрын

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.

Sir, for the second part, would I still use the same charge ratio that was originally derived?

@MichelvanBiezen

7 жыл бұрын

For the second part, Q inside is the total charge in the sphere since the whole sphere is no inside the Gaussian surface.

@Shhhchris

7 жыл бұрын

Got it. Thanks, sir. Greatly appreciated.

good sir g

Why the field inside the sphere isnt zero if the distribution of charge is symmetrical?

@MichelvanBiezen

5 жыл бұрын

Note that the charge is distributed all throughout the sphere, not just on the surface only.

i am confused about 5:30 what do you mean charge outside is the same as charge inside

@MichelvanBiezen

3 ай бұрын

The charge inside the Gaussian surface equals the charge inside the sphere (when r > R)

hello Sir, unfortunately i couldn't understand your aim in finding the ratio(in part where r

@MichelvanBiezen

7 жыл бұрын

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).

@rustamshahverdiyev4164

7 жыл бұрын

Michel van Biezen thank you sir )

how do we calculate the potential at the sphere center? What equation do we use?

@MichelvanBiezen

Жыл бұрын

V at the center = V at the surface + integral of E dr from the edge to the center.

@keithwallace5277

Жыл бұрын

@@MichelvanBiezen thank you!

Hello sir you never replied back from the specula relativity video for the KE rel Particle collision

@MichelvanBiezen

5 жыл бұрын

We try to answer some of the questions. We don't always have time to answer all.

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

@MichelvanBiezen

4 ай бұрын

The potential at the center of a sphere with uniform charge density is not zero.

@niall_al3059

4 ай бұрын

@@MichelvanBiezen so is the question faulty or something? Is it not possible to calculate the potential on the surface of the sphere?

@MichelvanBiezen

4 ай бұрын

Yes, just place the Gaussian surface right around the surface of the object.

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) ???

@MichelvanBiezen

9 жыл бұрын

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.)

@amartinez988

9 жыл бұрын

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. :)

@amartinez988

9 жыл бұрын

That's the example!

But why don't you have to find the ratios for when the Gaussian radius is larger than the object?

@MichelvanBiezen

5 жыл бұрын

Because then all of the charge is within the Gaussian surface.

so you only do V(gaussian surface)/V(sphere) when the radius of the gaussian surface is less than the object radius?

@MichelvanBiezen

5 жыл бұрын

Yes, and if the charge distribution is uniform within the sphere.

@user-pq1cj3hy3q

5 жыл бұрын

Michel van Biezen Much appreciated, thank you!

Sir pls. U have r

@MichelvanBiezen

Жыл бұрын

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.

@mondaks5774

Жыл бұрын

@@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

@MichelvanBiezen

Жыл бұрын

The method shown in the video is correct. If you get the same answer then your method is correct as well.

i don not understand the volume ratio part. plz clarify me.. Thankss for such´great videos by the way..

@MichelvanBiezen

9 жыл бұрын

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.

@zain-ul-aabdin3078

9 жыл бұрын

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....

@MichelvanBiezen

9 жыл бұрын

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

@zain-ul-aabdin3078

9 жыл бұрын

Michel van Biezen Million times thanks Sir....;))

Good lecture, but like charges repel each other. That's why high voltage power lines should be hollow! Kevlar inner core! ;) "skin effect"

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

@MichelvanBiezen

6 жыл бұрын

The answers are yes and yes. (If the sphere was a conductor, the excess charges would reside on the surface only)

@gaoxianglyu3450

6 жыл бұрын

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?

@gaoxianglyu3450

6 жыл бұрын

Because it is a conducting sphere. My mistake

@gaoxianglyu3450

6 жыл бұрын

Michel van Biezen this video belongs to insulating sphere or conducting sphere

@MichelvanBiezen

6 жыл бұрын

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.

k, but how do i find electric energy stored, possibly inside?

@MichelvanBiezen

Жыл бұрын

The energy stored will either be as electric fields or magnetic fields. It this case with stationary charges, it will be electric field

@tormentor2285

Жыл бұрын

@@MichelvanBiezen do i have to use the U=εE²τ/2? quite unusual not using it for a condensator

BUT the teacher told us electrical field=0 inside the conductor because the vectoe cancel each ather

@MichelvanBiezen

4 жыл бұрын

That is correct for a conductor. This is therefore not a conductor.

So, if this sphere is a conductor not insulator, and r is less than R , then E=0?

@MichelvanBiezen

5 жыл бұрын

Then all the charge would reside on the surface and you would be correct.

@asaadalalawi3561

5 жыл бұрын

@@MichelvanBiezen Thank you for your response and videos.

whats the difference between this and the other video (gauss law 3) they both seem to look for the electric field outside a sphere

@MichelvanBiezen

10 жыл бұрын

Susan, I updated the set at a later time and probably duplicated some.

@MichelvanBiezen

9 жыл бұрын

Hime Yoru Yes they are very similar.

@caido0913

9 жыл бұрын

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.

inside a sphere there is no charge so the electiric field inside the sphere wouldnt be equal to zero?

@MichelvanBiezen

6 жыл бұрын

If there is no charge inside the sphere, the electric field inside the sphere will be zero.

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.

@MichelvanBiezen

4 жыл бұрын

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.

@manmohanbisht69

4 жыл бұрын

@@MichelvanBiezen so when we talk about uniformly charged sphere it's not metallic. Since in case of metal charges reside on the surface..

@MichelvanBiezen

4 жыл бұрын

that is correct

good thumbnail choices

sir, i'd like to know the difference between part 3 and part 6... can you please explain

@MichelvanBiezen

7 жыл бұрын

In this video, the charge in the sphere is distributed throughout the sphere (not just at the surface only).

@elangz9201

7 жыл бұрын

Michel van Biezen thanks sir. youre the best

so this is an insulator ?

@MichelvanBiezen

8 жыл бұрын

+yassin alzaiem Yes, with a conductor all of the charges would be at the surface.

who is disliking the video :(

@tbvinicius

4 жыл бұрын

neighbors who don't like us, ex girlfriends LOL or kkkkkkk from brazil

I thought Gauss's Law was always in terms of surface area not volume?

@MichelvanBiezen

9 жыл бұрын

John, You are correct. (But you need the volume integral to calculate the charge inside the Gaussian surface.)

For the 1st one: Wouldn't all charges cancel due to symmetry?

@MichelvanBiezen

8 жыл бұрын

+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.

@oViTaLZzHD

8 жыл бұрын

Oh ok. Thanks Michael Michel van Biezen

Instruction unclear ended up getting a free amperemeter in my house now ,how?

why 4/3 pi r^2 and not just 4 pi r^3?

@MichelvanBiezen

10 жыл бұрын

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.

@HispanicMajority

10 жыл бұрын

Michel van Biezen Got it, thank you for clearing that up :)

If we were given the charge density could we have solved it using integrals?

@MichelvanBiezen

5 жыл бұрын

Yes, we have some videos showing how to do that.

why do we use r vs. R?

@MichelvanBiezen

9 жыл бұрын

Sana, No particular reason other than clarity. To indicate that there are two different radii.

@hypnotize8155

9 жыл бұрын

Got it! Can you please go over the properties of insulators and conductors and how they effect charge?

@MichelvanBiezen

9 жыл бұрын

Sana Sarfaraz There are some videos in playlist PHYSICS 39 which goes over dielectrics in capacitors.

Hello,professor do you have email.I have 3 problems. I want you to help me. Please

@MichelvanBiezen

6 жыл бұрын

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.;

@MichelvanBiezen

3 жыл бұрын

👍

You learn these things in college??,these are my 12th standard questions 🤧

@MichelvanBiezen

2 жыл бұрын

Where do you live?

@blkukreti4381

2 жыл бұрын

@@MichelvanBiezen in india sir,we don't actually have it in our books but teachers teach this because it come in exams

@MichelvanBiezen

2 жыл бұрын

Yes the high school curriculum is advanced in Indian high schools.

@blkukreti4381

2 жыл бұрын

@@MichelvanBiezen oh,thanks again for viedo sir

fank yu