Sakshi Shukla if there is a point charge at center and outside the shell, the total electric field would be the point charge plus the field outside of the shell
Sir a conducting spherical she'll having charge q uniformly distributed on its surface and Q charge is at the centre of given sphere. What is electrostatic potential at R/2 from centre?
First question is wrong. You just written '20 micro coulomb at the centre' and you are calculating as there is no charge. The question should be like "Hollow spherical conductor having charge of 20 micro coulomb"
You are right , it is likely that 20 micro coulomb must be the total charge of hollow spherical conductor, under this conditions total charge is uniformly distributed on outer surface of conductor so there is no charge inside conductor. Electric field is zero in this region. If a point charge were located at the center of an arbitary hollow conductor, electric field would not be zero in conductor because this charge would create electric field
So whenever it says "conductor" should we automatically think Gaussian surface? Also in the problem it says that there is a charge at the center. Is that just a typo?
Gauss' law enables us to calculate electric field in different regions containing high symmetry. It is valid for all closed surfaces and very useful in terms of mathematical process. So it can be used both conductors and nonconductors. According to Gauss' law outer charges do not contribute to find out electric flux and it is not important wherever charge is, providing that charge is inside the region
@Defend :er elektrik fiziği anlatan kitaplar pdfler ve ders notları son derece soyut ve akademik dil taşıdığı için bu derslerde öğrenciler oldukça zorlanmakta ve idrak edememektedir,bu da başarısızlığın temel nedeni olarak çıkar karşımıza
@@tempra87907 Merhaba bu videoda takıldığım bir noktayı sormak istiyorum size cevaplarsanız çok sevinirim. 2. kısımda sanki iç içe geçmiş iki hollow sphere varmış gibi davrandı yani yükü dağıtırken iç dediği kısım nasıl +8 i dağıttığı kısım olabilir? tam olarak algılayamadım. merkezden ilk çember kadarki kısım iç değil mi? buradaki durumu pek anlamadım ilk örnekte problem yok ama 2. karıştı yardımcı olursanız sevinirim
The electric field inside a conductor is always 0. This is because conductors have free-moving electrons, which will orient themselves to cancel the field created by the point charge. I know this comment is a year old but I hope this helps someone else!
I'm not gonna lie this was the first vid I've seen where I'm low-key confused. Nothing against your teaching, I just think this is a hard concept to begin with.
I had a problem where I needed to input the formula for finding the electric field at distance r from the center of two hollow sphere's containing a uniform charge of q-sub-1 and q-sub-2. Then, it asked what the charge was on the interior and exterior of the surface of the spheres.
In first example I understood that there is no electric flux if there is no charge inside the imaginary Gaussian surface. The cylinder is positively charged, which means that electic field is going outside of surface. So there should be electric field going towards hollow. But as per Gauss's law there is no electric. That is the confusing part. Why we have to take Gauss's law as granted ? Where is derivation?
If I am mistaken, please correct me. Bro, an electric field's movement is from positive charge to negative one. Nevertheless, there is just positive charge inside of the conductor, which means that electric field cannot exist there. Therefore, the electric field inside of the conducter is zero.
When you apply Gauss' law to calculate electric field in any region containing high symmetry,.In this law ''dA''factor called area vector always perpendicular to surface and outward, you must consider an imaginary Gaussian surface suitable for geometry. So in this question you are studying on conducting sphere and its area formula is also 4pir^2. To find out volume charge density you can use (Total charge/volume) rate.
there can be an external electric field coming from the charge clumped on the conductors surface! but when your gaussian surface is fully within the conductor and does not include the surface boundary, THEN the electric field is 0
@@roberttaylor4601 ok cool. Gauss’s law can be confusing, I’m just now understanding it only because I’m studying it for my midterm. Good luck with your studies!
@@DavidLopezcruz08 look up the ultimate gauss law review, it’s super helpful when it comes to answering those types of questions. Also WNY tutor has a playlist full of questions that he goes through. Goodluck you got this!
Sorry but its wrong .The Gauss' law is WRONGLY stated .. its not the product EA but the closed integral of the dot product of E and dA(or simply net FLUX) that is = Qenclosed/e. Surprised to see so many people took it at face value . And no , net charge inside a gaussian surface need not be zero for electric field to be zero . There could be electric field lines passing the gaussian surface but the net FLUX will be zero because net inward flux is gonna be equal to net outward flux ... Its simple physics.
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Correction: I meant to write that the +20uC charge should rest on the surface of the hollow sphere as oppose to the center.
The Organic Chemistry Tutor lmao I was about to say...
Holy crap! I was going nuts until I read this!!!
If charge is on center please tell me what the electric field be outside
Sakshi Shukla if there is a point charge at center and outside the shell, the total electric field would be the point charge plus the field outside of the shell
@@jdextlab OK thanks 😊
I was completely lost scrolling through videos, and this is the one that saved me. Thank you very much!
This was amazingly helpful. I woke up an hour early before my test hoping to find a youtube video to help and this was perfect.
Thanks for all the amazing help !! This is replacing my zoom class
Thanks iam very glad that we hv such a person to help we students when we get stucked up
IF I PASS MY MIDTERM, IT WILL BE BECAUSE OF YOU!!! THANK YOU
my man!!! this video came in clutch! thanks
Sir a conducting spherical she'll having charge q uniformly distributed on its surface and Q charge is at the centre of given sphere. What is electrostatic potential at R/2 from centre?
THANK YOU VERY MUCH FOR THE SECOND PART OF THIS VIDEO.IT HELP ME TO SOLVE A GREAT PROBLEM THAT I WAS FACING
i cant help but feel totally lost in the first 2 minutes. really wish you would redo this video for clarity.
I liked his second example much better
Nicely explained concept
First question is wrong.
You just written '20 micro coulomb at the centre' and you are calculating as there is no charge.
The question should be like "Hollow spherical conductor having charge of 20 micro coulomb"
I'm not exactly sure but I think its because an electric field inside a conductor according to Gauss's Law is considered to be 0.
You are right , it is likely that 20 micro coulomb must be the total charge of hollow spherical conductor, under this conditions total charge is uniformly distributed on outer surface of conductor so there is no charge inside conductor. Electric field is zero in this region. If a point charge were located at the center of an arbitary hollow conductor, electric field would not be zero in conductor because this charge would create electric field
lol that is exactly what I was thinking. You can't have a charge at the center of a conductor, yet that is exactly what the question says there is.
@@tempra87907 thanks guys, made me understand better
I like the way you solve the exercise
Wait, is the area with the yellow dotted line at 7:32 hollow or is the conductor the entire area from the second circle to the first circle?
It's everything. R goes from 0 to R2.
Thank you very much!! Wonderful teaching.
What would happen if the point charge inside and on the conductor had the same sign?
So whenever it says "conductor" should we automatically think Gaussian surface? Also in the problem it says that there is a charge at the center. Is that just a typo?
Gauss' law enables us to calculate electric field in different regions containing high symmetry. It is valid for all closed surfaces and very useful in terms of mathematical process. So it can be used both conductors and nonconductors. According to Gauss' law outer charges do not contribute to find out electric flux and it is not important wherever charge is, providing that charge is inside the region
@Defend :er teşekkür ederim mühendislik öğrencilerine calculus diferansiyel ve yüksek Fizik özel dersleri vermekteyim
@Defend :er elektrik fiziği anlatan kitaplar pdfler ve ders notları son derece soyut ve akademik dil taşıdığı için bu derslerde öğrenciler oldukça zorlanmakta ve idrak edememektedir,bu da başarısızlığın temel nedeni olarak çıkar karşımıza
@@tempra87907 Turkish much?
@@tempra87907 Merhaba bu videoda takıldığım bir noktayı sormak istiyorum size cevaplarsanız çok sevinirim. 2. kısımda sanki iç içe geçmiş iki hollow sphere varmış gibi davrandı yani yükü dağıtırken iç dediği kısım nasıl +8 i dağıttığı kısım olabilir? tam olarak algılayamadım. merkezden ilk çember kadarki kısım iç değil mi? buradaki durumu pek anlamadım ilk örnekte problem yok ama 2. karıştı yardımcı olursanız sevinirim
This video saved my life
You are a legend dude
How is e= 0 if there is a charge at the center? in the first example?
The electric field inside a conductor is always 0. This is because conductors have free-moving electrons, which will orient themselves to cancel the field created by the point charge. I know this comment is a year old but I hope this helps someone else!
Perfect video! What about r1=0 though? Do integrals come to play there?
Thank you❤
i hate physics but you're videos make it tolerable. thanks!
It says the charge is at the center. So how is E field at a distance 3m 0? 1:29
I would recommend the tutor to show a real picture of the speher in the upper corner to ease the imagination in the followers head.
Thank you very much.
cannot thank u enough :')
I'm not gonna lie this was the first vid I've seen where I'm low-key confused. Nothing against your teaching, I just think this is a hard concept to begin with.
U r the best men
bless your soul
I had a problem where I needed to input the formula for finding the electric field at distance r from the center of two hollow sphere's containing a uniform charge of q-sub-1 and q-sub-2. Then, it asked what the charge was on the interior and exterior of the surface of the spheres.
Thanks!
in the second problem:
when we are calculating the field at r1 why don't we calculate the field of the positive charge of the conductor
because r1
In first example I understood that there is no electric flux if there is no charge inside the imaginary Gaussian surface. The cylinder is positively charged, which means that electic field is going outside of surface. So there should be electric field going towards hollow. But as per Gauss's law there is no electric. That is the confusing part. Why we have to take Gauss's law as granted ? Where is derivation?
If I am mistaken, please correct me. Bro, an electric field's movement is from positive charge to negative one. Nevertheless, there is just positive charge inside of the conductor, which means that electric field cannot exist there. Therefore, the electric field inside of the conducter is zero.
thank u so much
you r great:)
but why E=0 inside there still electric feild
At second problem, is the spherical shell has 15 uc conductor or nonconductor? Could anyone give my any info about this?
Thankyou
thanks bro
you got a) wrong you said that there no charge at the center therefore E=0 but the question said +20 uC at the center
the question a it says that there is a positive charge at the center!
Thnq sir
WHY is it sometimes we use 4PIr^2(area) and sometimes 4/3PIr^3(volume)?
When you apply Gauss' law to calculate electric field in any region containing high symmetry,.In this law ''dA''factor called area vector always perpendicular to surface and outward, you must consider an imaginary Gaussian surface suitable for geometry. So in this question you are studying on conducting sphere and its area formula is also 4pir^2.
To find out volume charge density you can use (Total charge/volume) rate.
TX sir
I think the question was poorly worded
why can't we use 7 micro coulumbs for r2 electric field?
It is the charge enclosed within the Gaussian surface.
Matt is right, the 7 chatges are excluded from the surface with r2, they are in the surface r3-r2
someone can explain to me why the elecric field in the conductor is not just 0(it is one of the rules of conductors)?
there can be an external electric field coming from the charge clumped on the conductors surface! but when your gaussian surface is fully within the conductor and does not include the surface boundary, THEN the electric field is 0
on the last part what if both charges are positive/negative
the charges would get added up
Wouldnt r1 be equal to zero since its inside a conductor?
But there's a negative charge inside. Idk for sure but I'm pretty sure it's not zero.
@@DavidLopezcruz08 yeah I figured it out, if there’s a point charge inside then the q is equal to that
@@roberttaylor4601 ok cool. Gauss’s law can be confusing, I’m just now understanding it only because I’m studying it for my midterm. Good luck with your studies!
@@DavidLopezcruz08 look up the ultimate gauss law review, it’s super helpful when it comes to answering those types of questions. Also WNY tutor has a playlist full of questions that he goes through. Goodluck you got this!
Iweyo ndi dolo
Charge should be outside ri8?
it is spelt "right" not "ri8", stop being lazy.
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Bt i think in this electric field outside the sphere will be zero
It won't, because the hollow sphere isn't connected to a GND.
@@sutyi06 if charge is placed at the centre of the spherical conductor then what the electric field be outside the conductor??
Sorry but its wrong .The Gauss' law is WRONGLY stated .. its not the product EA but the closed integral of the dot product of E and dA(or simply net FLUX) that is = Qenclosed/e. Surprised to see so many people took it at face value .
And no , net charge inside a gaussian surface need not be zero for electric field to be zero . There could be electric field lines passing the gaussian surface but the net FLUX will be zero because net inward flux is gonna be equal to net outward flux ... Its simple physics.
Uhm,if there is no charge,then what is creating the electric field?
@@rachetmarvel931 the electric field can be created by an external charge(outside the gaussian surface)
thankss , made me understand better
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3 to 5 gold diggers a week!
sorry this is so so wrong
Nope you got it wrong man
Do not follow this, this is wrong
thanks bro