You sir, are a legend! I shall have to go back and like all these videos. I've been having so much fun that I forgot to like them. You have explained Gauss' Law and its applications so much better than my lecturer could ever hope to. Thank you so much! lasseviren1!!!
I dont think you will ever read my comment but I just want to thank you for all the helpful and clear videos! Now everything makes more sense than when my prof taught me. Thanks!
I realize that this is probably too late to be helpful, but in the previous video he told us that the shells were grounded, which would effectively cancel the positive charge on the outside of the sphere. He did not do that here, which means that the positive charges still exist and we need to take them into account when calculating the net charge. Hope that helped anyone who was confused.
Thank you so much for these videos. I finally have some idea of what's going on with Gauss's Law and how to derive the formulas using different surfaces!
I'm currently in an honors physics course with an awful professor, and I'd just like to let you know that your videos are very helpful. Thank you for making them.
No, it wouldn't be zero. That hollow insulating sphere would no longer have to have a charge of -Q on the inside wall so if you drew a Gaussian Sphere with a radius of r where b
Thank you so much! I think my grade went from a "d" to a "b". I have a test tomorrow and I was so nervous. I will tell you how I do tomorrow. Thanks for these videos! These are a life saver.
So in your first example, the system is grounded and thus the outer surface has a charge of zero. If it were not grounded, then the outer surface in that example would have a charge of +Q?
yes the outer surface will have a charge +Q as if it was earthed the electrons would come on the surface and the neutralize the charge so yea if its not earthed it will have a charge +Q on the outer surface
Im slightly confused with one matter how come's the first concentric sphere you did you said that E=0. However in this last concentric sphere you covered at the end of this video which seems to be the same amount of Q as the first concentric sphere but you said that E on the outside of teh sphere is equal to E= 2Q/4pieE0r^2. maybe i misunderstood something. Someone please help me =). thank you
sir in 6:42 how the charge b/w the area 2R and 3R is Q i.e. qnet=Q.. I THINK THAT as it is inside the metal so there is no electric field so i think that there is no electric charge.
-in the first exemple, the outer shell is not charged while in the second exemple the outer shell is charged with 1Q -in the first exemple we get the inner wall of the outer shell charged negative because we ground the outer shell hope this helps:)
In the first example, the outer sphere is grounded, so the charge the outside is 0. The total charge on the sphere is Q from the inner sphere + negative Q from the inside of the outer sphere. In example two, the outer sphere has net charge Q, so the net charge on the sphere is inner sphere (Q) + outer sphere (Q) =2Q
Charge is distributed on the surface. If it it was distributed in a full tennis ball, you'd have Q/Volume. Since this is equivalent to a baloon, it's Q/Surface.
What if we have another concentric insulator with no charge instead of the air do we have to count this q charge inside when we calculating between 2R -3R
I had the exact same example on a test, except that the charge on the nonconducting sphere = -Q. I got a question wrong that was asking about the electric field for r>c and the correct answer was -kQ/r^2 (r,hat) Anyone know why?
Because the charge on the nonconducting (outer sphere) is what is leaving the system as a whole; it radiates outward to everything around the sphere since that is the final charge leaving. If you are given the charge for the non-conducting sphere then there is no work that needs to be done just plug -Q into the equation to find E-fields (E=KQ/r^2) which gives you -KQ/r^2. In this example you are given the net charge and the charge on the inner sphere therefore you have to find the charge on the outer sphere but if you are simple given it the plug and chug. This makes since to me but my Teacher is very very good at visually explaining stuff so it might not for you. In that case I apologize.
+Naeem Hakimi hi dude. it's about charge density of an area, similarly, charge density of a length is lambda=Q/L and charge density of a volume is rho=Q/V.
You explain there is no flux inside the metal shell because there's no electric field there. You don't explain why there is no electric field there. You said you consider the inside sphere as a point charge with a field pointing radially outwards. Wouldn't there then be a electric field from that inside the outer sphere? Or is this cancelled due to the -Q
The outside shell is a conductor I think, and if so would mean all the charges are on the outside edges. No charge sits in the middle of a conductor, only insulators.
Its because total is net q. Since there is 3q + (-q) it equals to 2q. On the figure the spheres have +q charge anyway. Dont get confused with the charge distribution of the outer sphere. It has a total charge of +q.
it finally makes sense... I was like fuckkk gauss law says there is still an enclosed charge within the second sphere. Your video clarified how the field and charge can remain zero in the second sphere by modifying the outer charge of Q with the sum of the inner and outer charge on the second shell surface 2Q-Q = Q thank you.
You sir, are a legend! I shall have to go back and like all these videos. I've been having so much fun that I forgot to like them. You have explained Gauss' Law and its applications so much better than my lecturer could ever hope to. Thank you so much! lasseviren1!!!
Better than any other physics profs in my university,,
12 years later this man is still saving souls. thank you kind sir!
Lol. It makes me think of a remark I read elsewhere. It said I feel like Moses on the mountain for 40 days
Thank you so much for making these videos! 13 years later and you're still helping people! Liked all videos and subscribed.
Thanks man! 18 Years later and this video is still helping students. Simplest and yet most thorough explanation I've come across of this example
I dont think you will ever read my comment but I just want to thank you for all the helpful and clear videos! Now everything makes more sense than when my prof taught me. Thanks!
I realize that this is probably too late to be helpful, but in the previous video he told us that the shells were grounded, which would effectively cancel the positive charge on the outside of the sphere. He did not do that here, which means that the positive charges still exist and we need to take them into account when calculating the net charge.
Hope that helped anyone who was confused.
THANK YOU
Its never too late you saved me trouble 8 years later.
Thank you so much for these videos. I finally have some idea of what's going on with Gauss's Law and how to derive the formulas using different surfaces!
I'm currently in an honors physics course with an awful professor, and I'd just like to let you know that your videos are very helpful. Thank you for making them.
I FINALLY GET IT! Thank you! lol. Explaining the net charges using the gaussian surfaces in the metal helped me so much!
You make great videos, I appreciate your work and your devotion to help others!
what if the outer spherical shell was an insulator and not a conductor? Would the electric field at b
No, it wouldn't be zero. That hollow insulating sphere would no longer have to have a charge of -Q on the inside wall so if you drew a Gaussian Sphere with a radius of r where b
Thank you so much! I think my grade went from a "d" to a "b". I have a test tomorrow and I was so nervous. I will tell you how I do tomorrow. Thanks for these videos! These are a life saver.
Well? Are you going to tell us?
vASYL lEZUIK I think I did well but the teacher is going to release the results tomorrow. Will update.
vASYL lEZUIK I ended up with a C. It's not the best but next time I'll work harder and use these videos as supplement ealrier on.
i honestly belive i will ace my next test because of you
thank you
cleared a lot of my confusions. Thanks!
So in your first example, the system is grounded and thus the outer surface has a charge of zero. If it were not grounded, then the outer surface in that example would have a charge of +Q?
I hadn't thought about that, but that seems to make sense to me
I was stuck on that part. Thanks!
Yes
yes the outer surface will have a charge +Q as if it was earthed the electrons would come on the surface and the neutralize the charge so yea if its not earthed it will have a charge +Q on the outer surface
No it will have 2Q.....bcoz net is Q....if outer was Q and inner is Q than net will be 0
I am loving this! it makes Physics look easy!
cleared all my confusions Thank you so much;)
Im slightly confused with one matter how come's the first concentric sphere you did you said that E=0. However in this last concentric sphere you covered at the end of this video which seems to be the same amount of Q as the first concentric sphere but you said that E on the outside of teh sphere is equal to E= 2Q/4pieE0r^2. maybe i misunderstood something. Someone please help me =). thank you
I really like your videos. I have been recommending your website to my other physics students. Thanks for posting!
Question, for charge density, shouldn't we be using the volume of a sphere on the denom not 4pir^2?
You’re a saviour
sir in 6:42 how the charge b/w the area 2R and 3R is Q i.e. qnet=Q.. I THINK THAT as it is inside the metal so there is no electric field so i think that there is no electric charge.
-in the first exemple, the outer shell is not charged while in the second exemple the outer shell is charged with 1Q
-in the first exemple we get the inner wall of the outer shell charged negative because we ground the outer shell
hope this helps:)
In the first example, the outer sphere is grounded, so the charge the outside is 0. The total charge on the sphere is Q from the inner sphere + negative Q from the inside of the outer sphere. In example two, the outer sphere has net charge Q, so the net charge on the sphere is inner sphere (Q) + outer sphere (Q) =2Q
Area of a sphere is 4*pie*r^2? isnt that the surface area? Why not use Q/V instead?
Charge is distributed on the surface. If it it was distributed in a full tennis ball, you'd have Q/Volume.
Since this is equivalent to a baloon, it's Q/Surface.
What about electric field at distance 2R from center? Do we consider (-Q) or not?
what a great video, thanks so much :)
What if we have another concentric insulator with no charge instead of the air do we have to count this q charge inside when we calculating between 2R -3R
So helpful! Thank you very much
Awesome explanation
Not to confuse anyone but the outer charge has probably been added from somewhere.
Great help...Thanks a bunch
@Kuahana03 P.S Just to clarify im talking about E on the G.S outside of each concentric sphere
the "q.net", is that that charge of the field of that guassian circle. I get the math behind, but I'm confused of the terminology
very helpful ! thanks so much
Saved my life
I had the exact same example on a test, except that the charge on the nonconducting sphere = -Q.
I got a question wrong that was asking about the electric field for r>c and the correct answer was -kQ/r^2 (r,hat)
Anyone know why?
Because the charge on the nonconducting (outer sphere) is what is leaving the system as a whole; it radiates outward to everything around the sphere since that is the final charge leaving. If you are given the charge for the non-conducting sphere then there is no work that needs to be done just plug -Q into the equation to find E-fields (E=KQ/r^2) which gives you -KQ/r^2. In this example you are given the net charge and the charge on the inner sphere therefore you have to find the charge on the outer sphere but if you are simple given it the plug and chug. This makes since to me but my Teacher is very very good at visually explaining stuff so it might not for you. In that case I apologize.
sir, please explain about sigma =Q/A ? i think you never mention them in your videos
+Naeem Hakimi hi dude. it's about charge density of an area, similarly, charge density of a length is lambda=Q/L and charge density of a volume is rho=Q/V.
What exactly is qinner and what exactly is qouter?
THANK YOU
Thank you very much..... U legend
what is the charge density of the outer wall?
2q
You explain there is no flux inside the metal shell because there's no electric field there. You don't explain why there is no electric field there. You said you consider the inside sphere as a point charge with a field pointing radially outwards. Wouldn't there then be a electric field from that inside the outer sphere? Or is this cancelled due to the -Q
The outside shell is a conductor I think, and if so would mean all the charges are on the outside edges. No charge sits in the middle of a conductor, only insulators.
I don't understand how the total q enclosed was 2q...wouldn't it be 3q??
Its because total is net q. Since there is 3q + (-q) it equals to 2q. On the figure the spheres have +q charge anyway. Dont get confused with the charge distribution of the outer sphere. It has a total charge of +q.
The first sphere was grounded, so the charge on the outside of the sphere disappeared. Exam today, SO hope this comes up!
it finally makes sense... I was like fuckkk gauss law says there is still an enclosed charge within the second sphere. Your video clarified how the field and charge can remain zero in the second sphere by modifying the outer charge of Q with the sum of the inner and outer charge on the second shell surface 2Q-Q = Q thank you.
Good stuff, thank you!
Can u pls tell me that how it become inside shell as -Q pls rpy
thank you its awesome!!
Many thanks
I don't think outer charge is 2q it is against law of charge conservation
It's because Qnet is given as Q. Therefore fore charge to be conserved, the outer shell must have a density of 2Q
In the first case, the outer sphere was grounded. In the second case I don't think that sphere is grounded.
great!!!!
awesome
can any 1 tell about electric potential
Love from Duke University!
t-t-t-today, junior!!
by the way, this is not how you write Q.
this is how sigma is written, σ not like ꝺ or o'
@davidenelson how about you learn how to respect people when they're trying to help, damn it.