Physics 38 Electrical Potential (13 of 22) Potential Outside a Cylindrical Conductor

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Michel van Biezen

Michel van Biezen

Күн бұрын

Пікірлер: 51
@TG-to5nf
@TG-to5nf 7 жыл бұрын
Michael, you may just be responsible for Saving my Theoretical Physics Degree. You explanations are VERY good and examples varied.
@tongwu4742
@tongwu4742 8 жыл бұрын
Really really really helpful. I got all my midterms 100 after watching these tutorials. Thanks a lot!!!!!!!!!!!!!!!!
@ceyceyali
@ceyceyali 8 жыл бұрын
mr bizen, is the value of E corretct ? E = Rc*σ/Rg* ε in cylindiral conductor, in privious chapter you learn to me
@MichelvanBiezen
@MichelvanBiezen 8 жыл бұрын
+ceyceyali E is correct in the video. You can check by using Gauss's law.
@ceyceyali
@ceyceyali 8 жыл бұрын
Thanks mr. but cylinder like act line. why we do this ?
@MichelvanBiezen
@MichelvanBiezen 8 жыл бұрын
+ceyceyali It depends on how the charge distribution is given. From the outside of the cylinder the charge acts like a line charge. (It is the same with a spherical shell of charge, it acts like a point charge).
@lavenderwilliams1147
@lavenderwilliams1147 7 жыл бұрын
Why did we use kq/r^2 to find E on the spherical conductor but gauss's law on the cylindrical one
@MichelvanBiezen
@MichelvanBiezen 7 жыл бұрын
When outside the spherical conductor, we can think of it like a point charge. You cannot do that with a cylindrical conductor.
@lavenderwilliams1147
@lavenderwilliams1147 7 жыл бұрын
Michel van Biezen so I'm guessing we can treat the finite line charge as a point charge as well since you used kq/r^2 when determining the E but used gauss's law for the infinite line charge ? (vids 10/11). In the comment section of one of the videos, you said the reason we could not use gauss's law for the finite line charge was because the E was not perpendicular to the surface ? my question is how do we know that ? and how do we know when to use either methods to find E. Long question I know I'd appericate if you could clear it up for me, exams this friday
@walpurgoffnacht
@walpurgoffnacht 8 жыл бұрын
mr bizen, is the value of E is derrived using hte Gauss' law?
@MichelvanBiezen
@MichelvanBiezen 8 жыл бұрын
+Usamah Jundi E can be found using Gauss's law or by integrating over the charge distribution.
@walpurgoffnacht
@walpurgoffnacht 8 жыл бұрын
thank you sir
@akilakavisinghe7189
@akilakavisinghe7189 6 жыл бұрын
How did you determine the field using Gauss's law in this question?
@akilakavisinghe7189
@akilakavisinghe7189 6 жыл бұрын
Is there some approximation being done with the linear charge density? I thought we would be using the area charge density?
@akilakavisinghe7189
@akilakavisinghe7189 6 жыл бұрын
Oh I understand now sir the approximation is ok in this case. I was also wondering where I can videos on the derivation and application of the formula I = nqAvd (where vd is drift velocity)?
@rjaph842
@rjaph842 6 жыл бұрын
Akila Kavisinghe he approximated the charge distribution around the cylinder to be equal to that of a linear charge distribution?
@Ayush-og1nn
@Ayush-og1nn 5 жыл бұрын
Sir, whats the reason that electric potential can be assumed to be zero in cases of spheres but not in cylinder ??. Pleas give any intuitive explanation also
@Cyberspine
@Cyberspine 5 жыл бұрын
The cylindrical conductor is assumed to be infinitely long, whereas a sphere is only finite. On the other hand, in the cylindrical case the electric field is inversely proportional to the distance r, whereas in the spherical case it is inversely proportional to the square of the distance instead, r^2. You may know that 1/1 + 1/2 + 1/3 + ... is a divergent sum, whereas 1/1 + 1/2 + 1/4 + 1/8 + ... is convergent. So in the spherical case you will always get the delta V to be finite when compared with a point at an infinite distance, but in the cylindrical case this delta V will always be infinitely large.
@valeriereid2337
@valeriereid2337 Жыл бұрын
Thank you so very much for this excellent lecture. You are the best!
@MichelvanBiezen
@MichelvanBiezen Жыл бұрын
You're very welcome! Thank you for your positive feedback. 🙂
@omarsamy1144
@omarsamy1144 6 ай бұрын
why integrate from infinity?
@MichelvanBiezen
@MichelvanBiezen 6 ай бұрын
It is a technique ised with gravitational potential energy and with electrical potential energy, but it doesn't work with potential difference as is shown in the video.
@wongholeungwong5294
@wongholeungwong5294 7 жыл бұрын
I have a question why sometimes dV = Kdq/r. Sometime it is dv = E dr
@MichelvanBiezen
@MichelvanBiezen 7 жыл бұрын
It depends on the application. dV = E dr is always true.
@khaileng3020
@khaileng3020 3 жыл бұрын
Is the electric field given in the questions, or you already divide it , this question is use surface charge density or linear charge density??
@khaileng3020
@khaileng3020 3 жыл бұрын
Ok , i found out it is linear charge density
@jessmolen618
@jessmolen618 2 жыл бұрын
Thank you for the explaination!! Very helpful
@MichelvanBiezen
@MichelvanBiezen 2 жыл бұрын
You are welcome. Glad you found our videos. 🙂
@anchitjain4511
@anchitjain4511 7 жыл бұрын
sir biezen can you upload videos for variable volume charge density for spherical shells
@MichelvanBiezen
@MichelvanBiezen 7 жыл бұрын
Spherical shells cannot have volume charge density since they don't have volumes. But spheres do and here is an example: Physics - Gauss' Law (4 of 4) Variable Charge Distribution: Solid Sphere kzbin.info/www/bejne/pXa9in2hfKujodU
@anchitjain4511
@anchitjain4511 7 жыл бұрын
but the shells having inner as well as outer radius can?
@anchitjain4511
@anchitjain4511 7 жыл бұрын
and sir can u upload some videos on calculating total energy of a system of concentric shells
@MichelvanBiezen
@MichelvanBiezen 7 жыл бұрын
The charge on a shell is called surface charge density.
@anchitjain4511
@anchitjain4511 7 жыл бұрын
i am talking about a spherical non conductor with a spherical cavity centered at its centre. and now the charge that the sphere holds is between the sphere's circumference and circumference of the cavity.
@bull3asaur168
@bull3asaur168 4 жыл бұрын
Sir, Can I say ? While coming from infinity to A point moving direction is opposite to electric field so cos180=-1 so the formula must be -integral E.dr
@MichelvanBiezen
@MichelvanBiezen 4 жыл бұрын
The concept is correct. You can integrate from any starting point so that you will calculate the difference in potential.
@carbine000
@carbine000 9 жыл бұрын
Awesome video. Thank you!
@Irfan-vo6fh
@Irfan-vo6fh 4 жыл бұрын
Why here can't you just put the value of electric field you obtained before using gauss law
@Laxplanespotting-1
@Laxplanespotting-1 7 жыл бұрын
Mr van Biezen, I think the solution is wrong.because, the equation that you applied here suppose to be used for infinitely long straight wire. For the cylindrical conductor we have to apply another Gauss's law......
@MichelvanBiezen
@MichelvanBiezen 7 жыл бұрын
No, the video is correct. The approximation of an infinite wire is OK in this circumstance. This type of approximation is used often in many circumstances when appropriate.
@deshbandhu2007
@deshbandhu2007 4 жыл бұрын
Thanks a lot sir...
@MichelvanBiezen
@MichelvanBiezen 4 жыл бұрын
Most welcome
@rjaph842
@rjaph842 6 жыл бұрын
First am grateful for you uploads mr Michel.My worry is that electric field due to a cylindrical conductor is given by total charge(Q) thn the denomenator is right bt here you have used lampda instead of Q
@MichelvanBiezen
@MichelvanBiezen 6 жыл бұрын
The video is correct. Note that: The total charge = linear charge density x length
@rjaph842
@rjaph842 6 жыл бұрын
Michel van Biezen what of if i understand it this way:charge distribution on a cylindrical conductor is taken to be in the centre of the cylinder then the formula for electric field will assume that of a linear charge because the electric field due to a cylindrical conductor and a linear conductor are different
@mohsinahmad6078
@mohsinahmad6078 2 жыл бұрын
Just a video I want❤
@MichelvanBiezen
@MichelvanBiezen 2 жыл бұрын
Glad you found our videos. 🙂
@ElifArslan-l9g
@ElifArslan-l9g 3 жыл бұрын
thank you
@MichelvanBiezen
@MichelvanBiezen 3 жыл бұрын
You're welcome
@aliakashah3986
@aliakashah3986 3 жыл бұрын
thank you :)
@MichelvanBiezen
@MichelvanBiezen 3 жыл бұрын
You're welcome!
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