sirspamsalot2 Thank you for the comment. The memories of being a student and not understanding many things, and not knowing where to go to get help, encourage me to put these videos on line now that I am able to do so. That and the comments I get from you and many others.
+Michel van Biezen Dear Professor, how come we ignore the electric field produced by the +5 micro-coulomb charge on the inside of the ring? Wouldn't this produce an electric field?
+Jacob Cadwell Im going to ask the same question to my teacher next week in class. But if you just apply Gauss Law; then Qinside = -5. The +5 charge on the surface of the inner ring is not inside your Gauss surface. So thats the reason the +5 charge is ignored
what if we consider that the spherical conductor with cavity is a hollow non-conducting sphere and point cahrge 5Q is placed inside the hollw non-conducting sphere, what is the electric field for r < radious of inner hollow sphere R1 and the electric field for radious of inner hollow sphere R1 < r < radious of outor hollow sphere R2 ??
Alrighty boys, I got a 65 on my Calc III midterm and a bad grade on my internal mechanics midterm. This is my saving grace. I must nail the Physics II midterm.
For Physics, take a look at SBCC Physics or Matt Anderson Physics on youtube.. and for Calculus III I recommend looking up either Professor Leonard or Jhevon Smith (Smith teaches at my school, The City College of New York).. Good luck with your studies
+carlos caldera In general, if the charge is on the surface of a conductor, the electric field will be zero inside the conductor. (Q inside the Gaussian surface will be zero).
Why does an equal amount of positive charge as the negative charge enter the hollow area? Is it because the negative charge attracts an equal amount of positive charge by nature?
The "total net charge" carried by this entity would be +3uC? as you did (-5uC)+(5uC)+(3uC) for the largest gaussian sphere...? the +8uC you have written is just the total of positive charge carried by the entity?
1)If the conductor had a negative charge, then the electric filed inside wouldn´t be 0 (no matter what point) because there wouldn´t be a positive charge to "counter" the negative charge 2)If the material was NOT a conductor and had a positive charge, the electric filed could be 0 at some point - depending on the distribution of charge troughout the material and the distance away from the center - because then at some point there would be enough positive charge to "counter" the negative charge 3)If the material was NOT a conductor and had a negative charge, the electric filed would not be 0 (same case as 1) is this correct?
Professor Biezen I have this question where we have to find the electric field inside a sphere with a cavity on the side - the radius of the cavity is half the radius of the sphere - and then find the electric field at a point A which is a point on the surface of the big sphere and the end of the radius of the cavity, hope I'm clear..My professor said that for the first part we have to consider the electric field as if the cavity was not there, but for the second part of the question i have no clue. How do we solve this? Thanks in advance!
Sorry. I am not able to form a "picture" of the problem by the way you described it. Some details are missing such as is the sphere a conductor or an insulator, how much (and which type) of charge is placed on the sphere. Is the cavity centered at the sphere's center, or offset? Is the cavity entirely inside the sphere. Is there any charge inside the cavity. If yes, how much and of what type. etc.
@@MichelvanBiezen My mistake for not mentioning. So the sphere is an insulator and the volume charge density (rho) is constant. The cavity is inside the sphere but not at the center, on the right side of it - like imagine drawing a small circle in the middle but then shift it horizontally to the right half side of the big circle (sphere) until the end, and this small circle has a radius of R/2 (R being the radius of the big one). There is no charge inside the cavity. For the second question of this problem, we have to find the electric field at a point A that is the “tangent” point at the right end of the the whole figure where the cavity and the big sphere have 1 common point
It is now perfectly clear. The way to solve that problem is to first assume there is no cavity and find the electric field at that point. Then you assume the sphere is missing and you only have the cavity. Then you assume that the cavity is a sphere with the same charge density as the original sphere and caculate the electric field again at point A. Then subtract the second value from the first.
+Michel Van Biezen. Hi, I have a quick question. Why is E3 charge inside only -5 + 5 +3 and not -5 +5 +3 +8. R3 is pointing to the outside of the circle which includes the 8 microCoulomb?
- 5 uC of charge are placed in the inside object. + 8uC of charge are placed on the outside conductor. Thus 5 uC of charge migrate to the inside surface of the outside conductor and the remaining 3 uC of charge migrate to the outside surface of the outside conductor.
It depends on how the charge is distributed. If the charge only resides on the surface we use the area formula. If the charge is distributed throughout the sphere we use the volume formula.
Not sure if anyone has addressed this yet, i skimmed over the comments but didn't see it. At 5.05minutes, you wrote E2 4pi r2 and forgot to square r2. Doesn't make a difference because the E ended up being zero. But just an fyi. And you really rock! Thanks for sharing these videos with us!
wait... what does this mean for faraday cages? if the human inside has charge Q, the inside surface of the mesh will have -Q, shouldn't you be able to get a little bit of a zap touching the inside of the cage, after which you'd have exactly 0 charge?
cos(180) = -1 Therefore if you perform a dot product between 2 vectors (let's say A and B) and the angle between them is 180 degrees, the result will be negative AB
if the negative 5 microcoulombs of charge inside the cavity is made positive then where will the 5 microcoulombs inside the conductor reside? will they be repelled by 8 microcoulomb charge on the outside so as to stay in between these two charge distributions or is there no effect of the 8 microcoulomb charge on the charges inside the conductor?
Mr. Professor the net electric charge of atom is zero. How is it possible, if the strenght of positive electric field of nucleous diminishes with square of distance and the electrons which are further away accomplish this neutrality with their negativity? does this electric filed inequality produces Van der Waals forces?
Essentially for most atoms, the electrical charges are distributed in a non-symmetric fashion, which causes the molecule to be polar, which then in turn causes forces to exist between molecules, and these are known as the Van der Waals forces.
@@MichelvanBiezen yes I agree (dipol moment is not zero), but hydrocarbons attract hydocarbons (dipol moment is zero). Van der Waals forces have the lowest chemical binding energy and I do not know why and this fact keeps me awake at night. Thank You.
What if the charge in the center is greater than that of the conducting shell? For example -5uc in the center and total charge of +4 uc in the shell. Would the charges on the shell just try to arrange themselves to try and get as close to equilibrium as possible (so -1uc) on the inner surface of the shell so that if you create a gaussian surface with radius between the shell radii it would lead to -1uc charge? Or would the center charge induce charge +5uc from the shell leading to net field/charge still being zero inside any Gaussian surface created between the radii of the shell and outside the shell the field is like a point charge of -1uc?
Second shell theorem states: If a charged particle is located inside a shell of uniform charge, there is no net electrostatic force on the particle from the shell. Isn't this contradictory to what you tell us about E1 (not being zero)?
Why would you conclude that there is no force acting on the -5uC charge? However that said, it is correct to say that there is no NET force acting on the -5uC charge.
I have a question: Why didn't you use charge density to find the charge instead of using the total charge of the conductor? When does it matter if you are using charge density?
The assumption here is that the charge density is constant as a function of angle. Gauss's law doesn't require the knowledge of distribution as a function of radius, just how much charge is inside the Gaussian surface.
Since the conductor has a cavity, it has an inner surface and that inner surface has excess negative charge on its surface. Then when you draw a Gaussian surface surrounding that inner surface there will be charge inside the Gaussian surface and hence there will be an electric field. (Always better to start with the basic laws of physics, then to site generalities that require specific conditions for them to be correct).
Yes, it would cause a negative charge to collect on the inner surface of the outer ring and a positive charge on the outside surface of the outer ring.
Thanks for the reply, wanted to make sure I was following correctly :). But I have one additional question if you have the time. Does this occur because the negative charge attracts the closest protons? And this then causes somewhat of a chain reaction along the interior of the metal such that protons in the next infinitesly small rings alter their positions slightly in order to counteract the new negative charges, until those protons on the outer ring have been neutralized by the newly, correct me if I am wrong, exposed negative charges on the ring just before them?
The way it works is as follows: When there is an excess positive charge in the center, it will attract the negative charges towards the inner surface of the outer conductor, which leaves a shortage of negative charges on the outside surface of the outside conductor making that positive. With conductors, the excess charge always resides on the surface only.
wow, I am REVISITING this lecture.. and I have a question.. what IF the Inside charge was NOT at the center but instead was Halfway between the Center and the inner Surface?.. would the Charge on the Outside surface still be EVENLY Distributed?
If we are talking about an inside spherical conductor with charge (inside the outer conductor), it would be evenly distributed across its surface and so would the charge be evenly distributed on the outer spherical conductor with charge distributed across its inner and outer surface as shown in this video.
Hi Michel..... yeah, it's ME again... lol.... just curious here, if the charge INSIDE the sphere was a +5 C (rather than the -5 C you use in the video), ... and therefore the inner face of the cavity would be -5 C (rather than the +5 C in the video) ... and if the outside was still the +8 C... would that mean that the outside surface of the Sphere would be +13 C (that is 8+5 = 13 C) ? By the way, you used MicroCoulombs.. but either way....I couldn't find a MICRO symbol on my computer...
It depends on how many are packed onto the charged body. The more that are packed the more likely it becomes. Essentially they need something to jump to and they need a large enough potential difference (creating a large enough electric field) in order to jump across an air gap.
Hello, I have a question about the first gaussian surface. Since the gaussian surface is inside that cavity with charge -5uC, shouldn't the Qinside the gaussian surface be equal to a portion of the full charge in that cavity?
+Abdul Zahid The Q inside the Gaussian surface is always the total charge within the surface.(When you add positive and negative charge together they do cancel)
The first dashed line circle is smaller than 20cm (r1) so why did you give it a value of 20cm, when 20 cm is the radius of the first solid line circle. You said you were looking for the E.Field of the inner surface of the sphere but you have it a radius of 20cm, which that is coordinates of surface not the inner
No we were looking for the electric filed inside the inner surface (not of the inner surface). (Remember that you draw the Gaussian surface at the location you want to calculate the electric field.)
Since - 5 uC of charge is placed on the inner object and + 8 uC is place on the outer thick shell, 5 uC of the 8 uC will travel to the inside surface of the outer thick shell as that charge is attracted to the negative charge on the inner object.
@@MichelvanBiezen But then how will the electric field be zero inside the outer conductor? I mean, if we have +8 uC on the outer thick shell, then by Gauss's law: integral(E*dA) cannot be equal to zero since Qenclosed/ε is not equal to zero (but +8uC instead)...
i confused how to find the charge, pls verify my comment Some prof did not give the value of the charge, so i tried to find the charge by applying the formular. @E1, Q1-inside= charge density * A ( where A= 4pi R1^2) @E2, Q2-inside = Q1+Q2= (charge density-1)( 4pi R1^2)+(charge density-2)( 4pi R1^2) @E3, Q3-inside= Q1+Q2+Q3= (charge density-1)( 4pi R1^2)+(charge density-2)( 4pi R1^2)+(charge density-3)( 4/3 pi R3^3)
+Hamza Ali It would drive more of the positive charge of the sphere to the outside such that you would have negative charge on the inside of the hollow sphere.
Hello there, I have a question regarding E3. From what I understood, and also in your video (3 of 11) Spherical Charge, if the sphere is a conductor then Q for E3 should be Sigma times Area. However, in this video, Q remains just Q as if the sphere is uniformly charged. Would you elaborate, please? Also, thank you very much.
The electric field is always a function of the the total charge enclosed by the Gaussian surface. The method of calculating the total charge depends on how the charge is distributed.
Sir, Why is it that in a conductor, the charge is on its surface always and not spread uniformly. The Other Question is how is Electric field inside a conductor = 0 Please Help
The charges repel each other and thus want to be as far away from one another as possible. When they are on the surface, that geometric arrangement allows them to be as far away as possible.
Isn't E1 supposed to be negative since the E field line is in the opposite direction of the perpendicular A line? (180 degrees difference = cos180 = -1) so -EA...?
khelenU SInce the electric field is a vector quantity, it cannot be negative. All vectors have a magnitude and direction, and the magnitude of a vector must therefore always be positive. That said, the vector could be pointing in a negative direction (as defined). The electric field inside the cavity (E1) is pointing towards the center. If written as a vector in spherical coordinates it would be pointing inward (negative direction).
Sir i have a question. when u are calculating the electric field INSIDE the conductor shoulnt you take the ratio between (R2/R3) and multiply it to Q. I know it becomes zero because Q= -5+5=0 but i want to known if i have misunderstood what you said earlier. Thank you for your videos
Jean, I am not sure what you are asking. But the best way to answer is that according to Gauss' law, the electric field = zero if there is no net charge inside the Gaussian surface. I hope that helps.
a question please ?! what if there are a hollow conducting sphere inside a bigger hollow conducting sphera ?! what will be the sketch of the electric field with r ??!!!?!?!?
I thought that the electric field inside the conductor is always =0 , but if I am correctly following what you are doing there will be an electric field inside the conductor (the region with R2) it just cancels out in this one because the question used -5microC and +5microC. I AM CONFUSED.
Dayeed Khan, If you draw a Gaussian surface and there is charge inside, then the electric field CANNOT be zero. If you have a single conductor with charge on the outside and not charge inside the conductor, then yes, the electric field will be zero inside the conductor.
Michel van Biezen how would the field be inside the cavity if the charge enclosed in the cavity was +ve? where would the field lines be directed. I'm confused
Syed Karooney Remember that the electric field on a conductor (and non-conductor) depends on the charge INSIDE a Gaussian surface. If there is charge inside the conductor then there will be an electric field inside the conductor. If there is no charge inside the conductor, then there will be NO electric field inside the conductor
Where the hell is the 8MC charge ?! and the questions that are coming in our exam only tell you the inner charge and never mention the outer charge, and then they ask to calculate the E outside ! I still don't understand !
+Anas Khateeb There is a - 5 micro coulomb charge on the inner sphere. Then there is a + 8 micro coulomb charged placed on the outside sphere ( + 5 micro coulomb will migrate to the inner surface and + 3 micro coulomb will migrate to the outer surface.
+Michel van Biezen got it , but if the question tells you it's neutral shell and gives you the inner charge for example 5 mc,, that means on -5 mc on the inner surface, and 5 mc on the outer surface , isn't it ?!
Alejandro,
The 8 microcoulombs are distributed as 5 microcoulombs on the middle sphere and 3 microcoulombs on the outer sphere.
You are a hero, thank you for your videos. You truly are making a better world by providing these videos! Live long and prosper my friend!
sirspamsalot2 Thank you for the comment. The memories of being a student and not understanding many things, and not knowing where to go to get help, encourage me to put these videos on line now that I am able to do so. That and the comments I get from you and many others.
Dao Mahad
Yes the charge inside the inner Gaussian surface is - 5 micro-coulombs
The charge outside the Gaussian surface is ignored.
+Michel van Biezen Dear Professor, how come we ignore the electric field produced by the +5 micro-coulomb charge on the inside of the ring? Wouldn't this produce an electric field?
+Jacob Cadwell Im going to ask the same question to my teacher next week in class. But if you just apply Gauss Law; then Qinside = -5. The +5 charge on the surface of the inner ring is not inside your Gauss surface. So thats the reason the +5 charge is ignored
You sir, are my hero.
I thank Allah for making u a complete package........
Sir, you are amazing. Thank you so much for these videos. You saved me from a possible heart attack in the lecture.
what if we consider that the spherical conductor with cavity is a hollow non-conducting sphere and point cahrge 5Q is placed inside the hollw non-conducting sphere, what is the electric field for r < radious of inner hollow sphere R1 and the electric field for radious of inner hollow sphere R1 < r < radious of outor hollow sphere R2 ??
Wow fantastic sir. Now I can solve NEET PROBLEMS.
Alrighty boys, I got a 65 on my Calc III midterm and a bad grade on my internal mechanics midterm. This is my saving grace. I must nail the Physics II midterm.
For Physics, take a look at SBCC Physics or Matt Anderson Physics on youtube.. and for Calculus III I recommend looking up either Professor Leonard or Jhevon Smith (Smith teaches at my school, The City College of New York).. Good luck with your studies
I LOVE MICHEL VAN BIEZEN!!!!!!!!!!! I LOVE YOU AND AM INDEBTED TO YOU. I LOVE YOUR LECTURES!!!!!
Very useful, very well explained. Thanks a ton for posting this!
What would the electric field be if you were trying to find the electric field with a r < r1? Would it be 0 because it's within?
+carlos caldera
In general, if the charge is on the surface of a conductor, the electric field will be zero inside the conductor. (Q inside the Gaussian surface will be zero).
4:29 made my day
5:52, made you another day
Why does an equal amount of positive charge as the negative charge enter the hollow area? Is it because the negative charge attracts an equal amount of positive charge by nature?
That is correct.
thanx....Ur lectures really help alot ...indeed blessing for me
Superb explanation
indeed
you are a genius :)
he's not.
The "total net charge" carried by this entity would be +3uC? as you did (-5uC)+(5uC)+(3uC) for the largest gaussian sphere...? the +8uC you have written is just the total of positive charge carried by the entity?
1)If the conductor had a negative charge, then the electric filed inside wouldn´t be 0 (no matter what point) because there wouldn´t be a positive charge to "counter" the negative charge
2)If the material was NOT a conductor and had a positive charge, the electric filed could be 0 at some point - depending on the distribution of charge troughout the material and the distance away from the center - because then at some point there would be enough positive charge to "counter" the negative charge
3)If the material was NOT a conductor and had a negative charge, the electric filed would not be 0 (same case as 1)
is this correct?
Professor Biezen I have this question where we have to find the electric field inside a sphere with a cavity on the side - the radius of the cavity is half the radius of the sphere - and then find the electric field at a point A which is a point on the surface of the big sphere and the end of the radius of the cavity, hope I'm clear..My professor said that for the first part we have to consider the electric field as if the cavity was not there, but for the second part of the question i have no clue. How do we solve this? Thanks in advance!
Sorry. I am not able to form a "picture" of the problem by the way you described it. Some details are missing such as is the sphere a conductor or an insulator, how much (and which type) of charge is placed on the sphere. Is the cavity centered at the sphere's center, or offset? Is the cavity entirely inside the sphere. Is there any charge inside the cavity. If yes, how much and of what type. etc.
@@MichelvanBiezen My mistake for not mentioning. So the sphere is an insulator and the volume charge density (rho) is constant. The cavity is inside the sphere but not at the center, on the right side of it - like imagine drawing a small circle in the middle but then shift it horizontally to the right half side of the big circle (sphere) until the end, and this small circle has a radius of R/2 (R being the radius of the big one). There is no charge inside the cavity. For the second question of this problem, we have to find the electric field at a point A that is the “tangent” point at the right end of the the whole figure where the cavity and the big sphere have 1 common point
@@MichelvanBiezen is the problem still not clear? 😔
It is now perfectly clear. The way to solve that problem is to first assume there is no cavity and find the electric field at that point. Then you assume the sphere is missing and you only have the cavity. Then you assume that the cavity is a sphere with the same charge density as the original sphere and caculate the electric field again at point A. Then subtract the second value from the first.
@@MichelvanBiezen That makes a lot of sense! Thank you, thank you, thank you!!
+Michel Van Biezen. Hi, I have a quick question. Why is E3 charge inside only -5 + 5 +3 and not -5 +5 +3 +8. R3 is pointing to the outside of the circle which includes the 8 microCoulomb?
- 5 uC of charge are placed in the inside object. + 8uC of charge are placed on the outside conductor. Thus 5 uC of charge migrate to the inside surface of the outside conductor and the remaining 3 uC of charge migrate to the outside surface of the outside conductor.
WHY is it sometimes we use 4PIr^2(area) and sometimes 4/3PIr^3(volume)?
It depends on how the charge is distributed. If the charge only resides on the surface we use the area formula. If the charge is distributed throughout the sphere we use the volume formula.
THANK YOU SIR!
Excellent video!
Not sure if anyone has addressed this yet, i skimmed over the comments but didn't see it. At 5.05minutes, you wrote E2 4pi r2 and forgot to square r2. Doesn't make a difference because the E ended up being zero. But just an fyi. And you really rock! Thanks for sharing these videos with us!
Never mind, you fixed it at 6.25 minutes.
for finding electric field inside the conductor , we should take area = 4 pi (R2^2 - R1^2) ryt ??? , if not then help
wait... what does this mean for faraday cages? if the human inside has charge Q, the inside surface of the mesh will have -Q, shouldn't you be able to get a little bit of a zap touching the inside of the cage, after which you'd have exactly 0 charge?
Since these are spheres, you would have a hard time reaching the inside charge.
hello i don't understand why when E and A have cos 180 between them you removed the minus sign shouldn't we keep it ? and thank you
cos(180) = -1 Therefore if you perform a dot product between 2 vectors (let's say A and B) and the angle between them is 180 degrees, the result will be negative AB
if the negative 5 microcoulombs of charge inside the cavity is made positive then where will the 5 microcoulombs inside the conductor reside? will they be repelled by 8 microcoulomb charge on the outside so as to stay in between these two charge distributions or is there no effect of the 8 microcoulomb charge on the charges inside the conductor?
The you would have - 5 micro C on the inner surface and + 13 micro C on the outer surface.
Mr. Professor the net electric charge of atom is zero. How is it possible, if the strenght of positive electric field of nucleous diminishes with square of distance and the electrons which are further away accomplish this neutrality with their negativity? does this electric filed inequality produces Van der Waals forces?
Essentially for most atoms, the electrical charges are distributed in a non-symmetric fashion, which causes the molecule to be polar, which then in turn causes forces to exist between molecules, and these are known as the Van der Waals forces.
@@MichelvanBiezen yes I agree (dipol moment is not zero), but hydrocarbons attract hydocarbons (dipol moment is zero). Van der Waals forces have the lowest chemical binding energy and I do not know why and this fact keeps me awake at night.
Thank You.
What if the charge in the center is greater than that of the conducting shell? For example -5uc in the center and total charge of +4 uc in the shell. Would the charges on the shell just try to arrange themselves to try and get as close to equilibrium as possible (so -1uc) on the inner surface of the shell so that if you create a gaussian surface with radius between the shell radii it would lead to -1uc charge? Or would the center charge induce charge +5uc from the shell leading to net field/charge still being zero inside any Gaussian surface created between the radii of the shell and outside the shell the field is like a point charge of -1uc?
Second shell theorem states: If a charged particle is located inside a shell of uniform charge, there is no net electrostatic force on the particle from the shell.
Isn't this contradictory to what you tell us about E1 (not being zero)?
The statement is correct. (There is no INDUCED charge). But nothing prevents you from placing a charge there as was done in this example.
So there is no force working on the -5uC charge, despite the presence of electric field E1?
Why would you conclude that there is no force acting on the -5uC charge? However that said, it is correct to say that there is no NET force acting on the -5uC charge.
I got confused with force and net force! Thank you very much for your videos and quick responses!
I have a question: Why didn't you use charge density to find the charge instead of using the total charge of the conductor? When does it matter if you are using charge density?
The assumption here is that the charge density is constant as a function of angle. Gauss's law doesn't require the knowledge of distribution as a function of radius, just how much charge is inside the Gaussian surface.
For E1, i thought the electric field inside a spherical conductor would always be 0...i dont understand why put -5q if E=0
Since the conductor has a cavity, it has an inner surface and that inner surface has excess negative charge on its surface. Then when you draw a Gaussian surface surrounding that inner surface there will be charge inside the Gaussian surface and hence there will be an electric field. (Always better to start with the basic laws of physics, then to site generalities that require specific conditions for them to be correct).
Would a positive charge inside the cavity cause a negative charge to surround the inner cavity's surface?
Yes, it would cause a negative charge to collect on the inner surface of the outer ring and a positive charge on the outside surface of the outer ring.
Thanks for the reply, wanted to make sure I was following correctly :).
But I have one additional question if you have the time. Does this occur because the negative charge attracts the closest protons? And this then causes somewhat of a chain reaction along the interior of the metal such that protons in the next infinitesly small rings alter their positions slightly in order to counteract the new negative charges, until those protons on the outer ring have been neutralized by the newly, correct me if I am wrong, exposed negative charges on the ring just before them?
The way it works is as follows: When there is an excess positive charge in the center, it will attract the negative charges towards the inner surface of the outer conductor, which leaves a shortage of negative charges on the outside surface of the outside conductor making that positive. With conductors, the excess charge always resides on the surface only.
Thanks teacher for your help
you're amazing
wow, I am REVISITING this lecture.. and I have a question.. what IF the Inside charge was NOT at the center but instead was Halfway between the Center and the inner Surface?.. would the Charge on the Outside surface still be EVENLY Distributed?
If we are talking about an inside spherical conductor with charge (inside the outer conductor), it would be evenly distributed across its surface and so would the charge be evenly distributed on the outer spherical conductor with charge distributed across its inner and outer surface as shown in this video.
Hi Michel..... yeah, it's ME again... lol.... just curious here, if the charge INSIDE the sphere was a +5 C (rather than the -5 C you use in the video), ... and therefore the inner face of the cavity would be -5 C (rather than the +5 C in the video) ... and if the outside was still the +8 C... would that mean that the outside surface of the Sphere would be +13 C (that is 8+5 = 13 C) ? By the way, you used MicroCoulombs.. but either way....I couldn't find a MICRO symbol on my computer...
+Philip Y yes I think that would be correct if the inner sphere has +5 micro coulombs and the outer has +8 the net will be 13.
what about the 8 microcoulombs, that would make the q=11microcoulombs??
The 8 micro coulombs is the sum of the 5 and the 3 micro coulombs.
why electrons do not fly off the surfaces of a negatively charged body?
It depends on how many are packed onto the charged body. The more that are packed the more likely it becomes. Essentially they need something to jump to and they need a large enough potential difference (creating a large enough electric field) in order to jump across an air gap.
does this mean that total charge on surface of conductor is +8?
The total charge on both surfaces (inner and outer combined) yes.
Michel van Biezen oh so that is the reason why it wasn't added to the Qin; instead the 3microC was added :o
yeah if that's the case your final equation should be 8k/r^2 instead of 3k/r^2
For E3 why does Q=3 and not 8? Why does the 8 get ignored?
the enclosed charge is +8 = -5
Hello, I have a question about the first gaussian surface. Since the gaussian surface is inside that cavity with charge -5uC, shouldn't the Qinside the gaussian surface be equal to a portion of the full charge in that cavity?
+Abdul Zahid The Q inside the Gaussian surface is always the total charge within the surface.(When you add positive and negative charge together they do cancel)
but what if the charge on the outer sphere distributed uniformly ?
The charge is distributed uniformly.
For the first one why is R1 = 20cm, when the the gaussian surface is clearly evident from your drawing smaller than the area of the first inner shell.
R1 is the radius to the inner surface of the hollow sphere. The small Gaussian surface is inside the cavity of the hollow sphere.
The first dashed line circle is smaller than 20cm (r1) so why did you give it a value of 20cm, when 20 cm is the radius of the first solid line circle.
You said you were looking for the E.Field of the inner surface of the sphere but you have it a radius of 20cm, which that is coordinates of surface not the inner
No we were looking for the electric filed inside the inner surface (not of the inner surface). (Remember that you draw the Gaussian surface at the location you want to calculate the electric field.)
whats the electric field along the inner surface, when R= 20 cm
Set R1 equal to 20 cm (0.2 m)
Dear sir, what if there was no 8 coulombs outside ? what if its just free space ?
If no charge was added to the outside conductor, there would be a charge of + 5 uC on the inside surface and - 5 uC on the outside surface
Michel van Biezen thank you so much sir, your videos are the best out there
How do you know that Qenc for E2 is (-5uC + 5uC) ?
Since - 5 uC of charge is placed on the inner object and + 8 uC is place on the outer thick shell, 5 uC of the 8 uC will travel to the inside surface of the outer thick shell as that charge is attracted to the negative charge on the inner object.
the charge on the outside has to move to the inside of the sphere because of the force exered by the neg charge
@@MichelvanBiezen But then how will the electric field be zero inside the outer conductor? I mean, if we have +8 uC on the outer thick shell, then by Gauss's law: integral(E*dA) cannot be equal to zero since Qenclosed/ε is not equal to zero (but +8uC instead)...
i confused how to find the charge, pls verify my comment
Some prof did not give the value of the charge, so i tried to find the charge by applying the formular.
@E1, Q1-inside= charge density * A ( where A= 4pi R1^2)
@E2, Q2-inside = Q1+Q2= (charge density-1)( 4pi R1^2)+(charge density-2)( 4pi R1^2)
@E3, Q3-inside= Q1+Q2+Q3= (charge density-1)( 4pi R1^2)+(charge density-2)( 4pi R1^2)+(charge density-3)( 4/3 pi R3^3)
What would be the effect on E2, if a positive charge was placed in the cavity of the sphere.
+Hamza Ali It would drive more of the positive charge of the sphere to the outside such that you would have negative charge on the inside of the hollow sphere.
You are a beast, thank you so much!
Hello there,
I have a question regarding E3. From what I understood, and also in your video (3 of 11) Spherical Charge, if the sphere is a conductor then Q for E3 should be Sigma times Area. However, in this video, Q remains just Q as if the sphere is uniformly charged.
Would you elaborate, please?
Also, thank you very much.
The electric field is always a function of the the total charge enclosed by the Gaussian surface. The method of calculating the total charge depends on how the charge is distributed.
@@MichelvanBiezen , so why exactly would you disregard the 4(pi)r^2 for Qinside. In this video Q = sigma or Q = sum of sigmas
thank u kindly sir
Sir, Why is it that in a conductor, the charge is on its surface always and not spread uniformly.
The Other Question is how is Electric field inside a conductor = 0 Please Help
The charges repel each other and thus want to be as far away from one another as possible. When they are on the surface, that geometric arrangement allows them to be as far away as possible.
Thanks a lot
Isn't E1 supposed to be negative since the E field line is in the opposite direction of the perpendicular A line? (180 degrees difference = cos180 = -1) so -EA...?
khelenU
SInce the electric field is a vector quantity, it cannot be negative.
All vectors have a magnitude and direction, and the magnitude of a vector must therefore always be positive.
That said, the vector could be pointing in a negative direction (as defined).
The electric field inside the cavity (E1) is pointing towards the center.
If written as a vector in spherical coordinates it would be pointing inward (negative direction).
Thank you for the video but what is the 8 micro coulombs indicating?
It indicates the total charge on the sphere. I think you are waiting for the results of your physics exam btw.
Sir i have a question. when u are calculating the electric field INSIDE the conductor shoulnt you take the ratio between (R2/R3) and multiply it to Q. I know it becomes zero because Q= -5+5=0 but i want to known if i have misunderstood what you said earlier.
Thank you for your videos
Remember that with conductors, the charge resides on the surface only.
Please help. why An object is enclosed within a metallic spherical shell having surface charge + q C. The object will carry charge -q C.
This is just an example in order to help us understand the concept of Gauss's law.
If the net charge wasn't 0, it would mean a Eletric field not = 0 in the surface of a condutor sphere?
Jean,
I am not sure what you are asking.
But the best way to answer is that according to Gauss' law, the electric field = zero if there is no net charge inside the Gaussian surface.
I hope that helps.
a question please ?!
what if there are a hollow conducting sphere inside a bigger hollow conducting sphera ?!
what will be the sketch of the electric field with r ??!!!?!?!?
It depends on how much extra charge exists on the inner sphere and the outer sphere, as they do interact.
amazing, thank you so much
Lifesaver! Thanks!
I thought that the electric field inside the conductor is always =0 , but if I am correctly following what you are doing there will be an electric field inside the conductor (the region with R2) it just cancels out in this one because the question used -5microC and +5microC. I AM CONFUSED.
Dayeed Khan,
If you draw a Gaussian surface and there is charge inside, then the electric field CANNOT be zero.
If you have a single conductor with charge on the outside and not charge inside the conductor, then yes, the electric field will be zero inside the conductor.
Oh yes I get it now. Thanks =)
Michel van Biezen how would the field be inside the cavity if the charge enclosed in the cavity was +ve? where would the field lines be directed. I'm confused
Syed Karooney and also if the sphere was non-conducting
Syed Karooney
Remember that the electric field on a conductor (and non-conductor) depends on the charge INSIDE a Gaussian surface.
If there is charge inside the conductor then there will be an electric field inside the conductor. If there is no charge inside the conductor, then there will be NO electric field inside the conductor
Thanks sir
I hope you solve it with no charge given, so you have to make integral
Thank you!
I feel educated
LEGEND
What if I need to find Electric potential?
what if the charge or the cavity is not at the centre of the sphere
Then it would be a much more difficult problem.
yes but can it be answered qualitatively like what would be the effect on the field at different places
what if the cavity charge is positive.what changes?
Ricz S
Then all of the charge on the outer ring would reside on the outside surface.
+Michel van Biezen And if that happened, the electric field between the cavity and the surface would not be 0, right?
Sir, what is the cavity?
The spherical conductor is hollow. The hollow part is the cavity and it contains a 5 uC point charge.
PERFECTO!!
Where the hell is the 8MC charge ?!
and the questions that are coming in our exam only tell you the inner charge and never mention the outer charge, and then they ask to calculate the E outside ! I still don't understand !
+Anas Khateeb There is a - 5 micro coulomb charge on the inner sphere. Then there is a + 8 micro coulomb charged placed on the outside sphere ( + 5 micro coulomb will migrate to the inner surface and + 3 micro coulomb will migrate to the outer surface.
+Michel van Biezen got it , but if the question tells you it's neutral shell and gives you the inner charge for example 5 mc,, that means on -5 mc on the inner surface, and 5 mc on the outer surface , isn't it ?!
+Anas Khateeb That is correct.
I love u men
lecturer with a bow tie, how stereotypical can you get?
ma mind blown
Sup bruh
👍