Physics 39 Capacitors (10 of 37) The Spherical Capacitor
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- Опубликовано: 15 янв 2025
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In this video I will develop the general equation for capacitance of a spherical capacitor.
Next video can be seen at:
• Physics 39 Capacitor...
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Tushar,
SInce R2 is larger than R1, R1 - R2 is a negative number.
Capacitance cannot be negative.
Rather than working it out as a strict mathematical problem, and ignoring the physical properties will leave you guessing (or you have to memorize the various rules when integrating over the electric field to find the change in voltage.
Better just to look at the problem and determine what sign it needs to be.
(That is what they did to set up the rules for the sign)
It's been long (almost since when I was in grade 10) since when I started following this channel and I'm still coming back to it even though I'm currently in 3rd year doing Math-Phys major. Just wanted to say thanks to Michel for creating such clear and concise videos on physics. Wishing your channel getting more views by the day :)
Thank you for sharing. It is good to see that the videos are helping math and physics students around the world.
indeed
youre becoming my best youtube teacher...
This is an excellent lecture, thanks for not making assumptions about what students should know, in that you have not missed a step, the physics and the math just come together beautifully. As someone who will some day teach physics and and mathematics I can only hope that I will be able to help my students understand the way you have helped me to understand.
you are welcome. By what I can tell from your comment, you will be an excellent teacher.
Thanks ......
First time i understand this topic through your lecture.
Thank you very much Sir..........
super helpful, I had no idea why some textbooks randomly flipped the limits of integration but now I know it’s to make delta V positive!
Glad it helped!
"What do we say to the professor kids?"
"Thanks for saving out A**es professor."
Mechanics was a joke compared to E M
+Максим «Сергеевич» Пушкаш Mechanics is way easy but it still got its perks
That's because in Mechanics you can easily visualize the scenario and write an equation that follows. In E&M, you can't visualize electrons flowing in an electric field and are only guided by the given equations.
Did you get this problem from a particular textbook? If so which one
Hey man! What mic did you use for the video? Thanks I love Physics
Just a cheap $20 in-line one.
Highly appreciated your effort sir. It was a wonderful lecture. Thank you so much
i had to say alot but kinda running short on time,
so i'll just say thank u sir, god grants u success
Thank you. God bless you as well.
Wow....so good ...it helped me to understand a big question .. thanks
Hello Professor, when you switched R2 and R1 shouldn't you have erased the negative sign on the first integral?
The sign of the delta V depends on the direction of the electric field and the direction of travel only. If you travel in the direction of the electric field, the potential decreases and if you travel in the opposite direction, then the potential increases
Sir the video was very Good and concise But in my reference book i saw thath the outer shell was earthed and capacitance was calculted then in next case the innner shell was earthed and capacitance was calculated but here I wanted to ask can we calculate the capacitance without earthing.
The only thing that matters is the potential difference across the plates. Grounding one side or the other side, simply places a reference voltage of 0 V on that side. As long as the voltage difference between the plates is not affected it doesn't matter.
Tx man for the lecture
You are welcome. :)
Alright
v=-(integral)E.dr
here angle between E and dr is 180deg
so the capacitance comes out to be c=R1R2/k(R1-R2)
please correct me where am i wrong????
I have got the same problem. The integral should be from R2 to R1, but without the minus sign. (Because *E* • *dr* is negative)
In Concepts of Physics by H.C. Verma, the same integral is given (which Michael mentioned), but I can't understand.
As your comment is 4 years old, you have already completed 12th. Currently I'm in 10th. Can you explain me why there is a minus sign?
@@1_adityasingh HC Verma's this particular derivation is not technically correct. Find the integral from smaller to bigger radius and you will get your answer (also realise that in doing so you are finding potential difference of negative sphere minus positive sphere, so remember to multiply both sides with negative in later stage).
Thank you for the explanation and example ^^
You're very welcome! 🙂
I have one doubt, why didn’t the presence of positive Q charge on the inner shell induce a -Q charge on the inner surface of the outer shell, since the outer shell is a conductor and field inside it must be zero
There is always an electric field outside a distribution of charge (as in the inner ring), therfore there must be an electric field between the inner sphere and the outer sphere
Why the outer shere 's charge didn't get attract by inner charge
thank you so much this was so helpful
Glad it helped!
Why is the electric field not determined by the charge on the outer sphere as well?
The electric field only acts outwards, thus the charge on the outside of the sphere does not affect the electric field inside the sphere.
@@MichelvanBiezen Thank you for your reply. But why doesn't the charges on the boundary of the outer sphere produce an electric field radially outward everywhere as expected, including in the interior as well?
@@josephghobrial5985 This is a capacitor. Electric field outside -Q shell is zero. Why? Simply, apply Gauss’s Law. Qin = Q - Q = 0
Hi! Thanks for your videos. I think I have a little of confusion . Isn't the electric field inside a conductor zero ? Then why is there between R1 and R2 an electric field? Thank you a lot!
Gauss's law stated the the electric field strength multiplied by the area of the Gaussian surface is proportional to the charge enclosed by the Gaussian surface. So bottom line, there will always be an electric field around an enclosed charge.
What about the potential at R2. How much is that?
We don't reference the potential AT A PARTICULAR LOCATION, we only calculate the potential DIFFERENCE BETWEEN TWO LOCATIONS.
Is this the case when inner sphere is earthed and Outer sphere is charged?... and there exists some other formula for the case when the inner sphere is charged and the outer sphere is earthed?
This is in response to your other question. Since the capacitors are no longer connected, the charge on each of them must remain the same. In the case of the parallel capacitor, the electric field between the plates is a function of charge density only, thus the electric field must remain the same. Since the potential between the plates = E x d , the potential must decrease due to the decreased distance. I would expect the effect on the spherical and cylindrical capacitors to be the same since the charges which are attracting each other are being pulled apart, which would require work, which would add energy to the system which would require the potential to go up (since the charge remains the same).
Which question sir?
Sorry because I am very confused up...
In tuition I am being taught magnetism I myself am practicing current electricity whereas I completed capacitor a few days ago and need to keep it on revision
It was a wonderful video and i understood it very well.. Tqsm😇☺🙏🙏
Glad it was helpful. 🙂
I think the title of the lecture should be edited to indicate that this lecture number "(10 of 37) ", there appears to be two (1 of 37) in this series.
Why don't we account for the outter shell when calculating the electric field?
+Терорист Талибан
The strength of the electric field between the inner and outer shell only depends on the charge on the inner shell, (see Gauss's law).
+Michel van Biezen I was also wondering this , when R=R2 wouldn't we then have to factor in the outer shell?
because if that outer shell didn't have any charge inside it the electric field anywhere inside it would be zero since charge is uniformly distributed. So moving around inside of it, where the field is always zero, means the outer shell doesn't really affect the electric field.
Терорист Талибан
Electrons from Vbatt induces a field at the inner sphere which induces current into the outer
If planets such as Earth have a conductive and ferromagnetic spherical nickle/iron core, surrounded by an insulating rock mantle, and then it surrounded by a conducting crust and atmosphere, would this not constitute a spherical RLC, (Resistor Inductor Capacitor), circuit? Also, if there are "Birkeland currents", (Google or RUclips the term if you are not familiar with it), or what NASA refers to as "magnetic ropes", connecting the poles of the Earth and other planets in our solar system to the poles of the Sun, would not each planet connected to the Sun in this manor act as an RLC network load? The resistance would mainly come from the overall pole to pole resistance of the crust and atmosphere. The inductance would come from the large ferromagnetic nickle/iron core mass. And the capacitance would come from the conductive core, inside an insulating mantle, inside a conductive crust and atmosphere. If these types of concepts are new to you, I would recommend Googling and RUclipsing the phrase, "Electric Universe". Here's as good a place as any to start... ruclips.net/video/5AUA7XS0TvA/видео.html
Sir in most of the tutorial they show charge on inner surface of outer shell but u showed it outer side...why is that so...please reply
On this video there is charge on the inside surface and the outside surface. With conductors excess charge will only reside on the outside surface.
Thank you very good course
I think you should change the title number (1 of 37) to (10 of 37)
Thank you for pointing that out. We changed it.
It is very wonderful👍
Thank you. Glad you found our videos. 🙂
How I wish i have my high school physics teacher by my side. He's so good at teaching me these stuffs.
thank you so much it was perfect.
Nice class
Thank u so much sir ❤️❤️
Most welcome
thank you michael van biezen
You are welcome. Glad it helped. 🙂
You switched the boundaries of the integral but the sign didn’t changed. Why?
Where exactly are you referring to? (Note that the voltage increase when traveling in the opposite direction to the direction of the electric field)
@@MichelvanBiezen At 3:19, the *E* field is opposite to *dr* i.e. the angle between *E* and *dr* is π. So *E* • *dr* should be - *E dr* and the minus sign should be removed from the front of the integral sign, right?
@@1_adityasingh dr direction is always pointing outward whether you switched the limit of the integral or not. i think you are confused because when you reverse the limit of an integral in Calculus class, you have to switch the signs. Here is a different story. Here, your signs depends on the direction of Electric Field and dr. Since dr is 0 at the center and positive at R1 and R2 its direction is outward. You can always play around with the integral and if you move from Low Potential to High Potential and get Negative sign, you have to know that there is a problem in your integral setting and you can correct it easily.
thank you
Most welcome
If I understood correctly and I would be more than happy that someone will correct me in case that I am not- the reason he keeps the negative sign is because the potential difference in R2 is smaller than that of R1 so when integrating from R2 to R1 we are getting a negative result- hence he needs to put a negative 1 to cancel that minus.
You are correct. 🙂
You are a god, sir!
Not in the least.
Michel van Biezen there are very people who does so much for the better understanding of their students, and with such a welcoming attitude. You are definitely one of em. keep helping us sir :)
Sir I am confused ,how these shells acquire this charge.
In my textbook ,earthing is also include but I can't understand it.
Can you explain me about it ,,,,,please.....
The purpose of the problem is to learn how capacitors work and how to calculate the capacitance, not how the charge is placed on the plates. However that said, if you ground one of the plates and then apply a voltage to the other plate you will place charges on the plates.
Very nice explaination
Thanks and welcome
sir,what confuses me is E is pointing radially outward but the path is chosen inward.so E.dr=-Edr shouldnt it be the case??
Since the change in voltage = E dr The voltage change is negative traveling in one direction and positive traveling on the opposite direction
When you change limits doesn't *E.dr* changes signs? (you never took the integrand as vector dot product, so technically this derivation is wrong although you get correct result somehow)
@Debanjan Bhattacharjee You think 'dr' is radially outwards?? 2:33
sir i think the -q charge should be in inside boundary of the outer spherical shell????
This question is in a textbook as written. Why do you think the negative charge would be inside the shell?
You can put the -Q inside and +Q outside, but you have to notice that (the dot product) E . dr = -E dr because electric field direction now inward which is opposite to dr direction. The negative sign of the integral will be cancelled. You can now integrate from R1 (low potential) to R2 (hight potential) and you will arrive to the same answer as the Professor.
Thank you !
Why isn't E constant? The separation between the spheres' surfaces is constant.
The electric field spreads out over an increasing volume, thus it is expected to be smaller farther out (as the electric field lines spread out).
why didn't the sign change as we change the limits as now we have increasing potential
+rahul SHAILY
When calculating the potential the sign is determined by definition, so as to ensure that the potential decreases when moving away from positive charges and the potential increases when moving away from negative charges.
No, by definition the sign is negative. Even switching the two points you don't need to change the sign, but just take into account that actually you are integrating the scalar product between E and dR, so you can just use a different angle between the two vectors (theta = 0 if you are integrating in dR from R1 to R2, and theta = pi integrating in dR from R2 to R1)
Liked your lecture and bow tie xd
You use incorrect relation between E and V. Since E is a conservative vector field the correct relation is: V (B) -V (A) = - definite integral (from A to B) from dot product E * dr. In your calculation, to get a good result you’ve had to change the interval of the integral. Using the correct relation, nothing has to be changed or determined.
I WAS RIGHT....MECHANICS IS EASIER !!!!