late but for charge enclosed it depends on if the object is a conductor (in which charge is located on surface) or if it is a nonconductor(when it is spread out). You will always use surface area for dA however
Because its a hollow conductor inside the shell there is charge but its hollow and its infinitely little thick (I propably made a syntax mistake hereXD)
if you have a Gauss reading of 1200Wb/m^2 and a sphere with a radius of 2.5mm. Could you find the magnetic flux of the bead? How would you determine the magnetic flux?
give me solution pls:a spherical conductor radius 0.1 is charged uniformly. charge on the surface is 16π c calculate the a) surface charge density. 2) electric field on the surface and 3)electric field at a distance 0.05m from the center of the sphere.
surface charge density is just 16 pi divided by the surface area of the sphere. The electric field on the surface is determined using gauss's law and the the field inside a conductor is always Zero
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Best teacher of physics man.
You are an amazing teacher! I love how you explain EVERYTHING, even the little details. Thank you so much!
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WHY is it sometimes we use 4PIr^2(area) and sometimes 4/3PIr^3(volume)?
late but for charge enclosed it depends on if the object is a conductor (in which charge is located on surface) or if it is a nonconductor(when it is spread out). You will always use surface area for dA however
Because its a hollow conductor inside the shell there is charge but its hollow and its infinitely little thick (I propably made a syntax mistake hereXD)
if you have a Gauss reading of 1200Wb/m^2 and a sphere with a radius of 2.5mm. Could you find the magnetic flux of the bead? How would you determine the magnetic flux?
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@@jimhalpert9803 lmfao
why can you use the same Q value for part b if the radius is different?
charge is a constant regardless of radius, E field will vary with radius
Your are the best...
Can I ask you 2 spherical gaussian problems? I have answer's key.
Look, if the surface of the spherical conductor is considered the charged point then why taking r, radius from the center of the charged sphere
give me solution pls:a spherical conductor radius 0.1 is charged uniformly. charge on the surface is 16π c calculate the a) surface charge density. 2) electric field on the surface and 3)electric field at a distance 0.05m from the center of the sphere.
surface charge density is just 16 pi divided by the surface area of the sphere. The electric field on the surface is determined using gauss's law and the the field inside a conductor is always Zero
you are the best
thanks
Fix ur quality jeez. the 144p chemistry tutor
I think 360 isn't really that bad?
This man really wants to see white text on black background in 1080p