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Systems Interactions and Changes Videos 8 videos

AP Physics 2: 1.2 Systems Interactions and Changes
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AP Physics 2: 1.1 Systems Interactions and Changes
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AP Physics 2: 2.3 Systems Interactions and Changes
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AP Physics 2: 2.3 Systems Interactions and Changes. Calculate the emf induced across the wings of the starship.

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AP Physics 2: 2.4 Systems Interactions and Changes 3 Views


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Description:

AP Physics 2: 2.4 Systems Interactions and Changes: Determine the radius R1.

Language:
English Language

Transcript

00:04

And here's your shmoop du jour brought to you by electrical

00:06

entertainment rumor has it that there are ways to be entertained that don't [man sitting on a couch watching television]

00:10

involve electricity but that involves scary things like activities that are

00:15

done outside ooh sounds gross we'll stick with cat videos all right, sparky

00:20

the shocker is the star electrical entertainer act of the 1730's parties [Sparky on stage holding a positive and negative balls]

00:25

collected a charge of eight coulombs on a metal sphere of a radius R sub 1 he

00:30

has an out-of-town gig and wants to take some of the charge with him he connects

00:34

his sphere to another sphere with a thin conducting wire the second sphere has a

00:39

radius of 12 centimeters and its uncharged before the connection take a

00:43

look the following diagram right here now once the system has reached [diagram of two spheres]

00:47

equilibrium the charge on sphere 2 is fixed Coulomb determine the radius R of

00:52

the first sphere and here the potential answers... Okay well in the 1730's this was [woman using her mouth to throw a chair in the air]

00:58

cutting-edge entertainment never been so glad to not live in 1730's but we're sure

01:03

seeing electricity in action for the first time back then was well pretty mind [man staring at a blank TV screen]

01:07

blowing. When the two spheres reach equilibrium the surfaces of each will

01:11

have the same electric potential at a spheres surface and farther out the

01:16

electric potential will be equivalent to that of a point charge located at the [finger pointing to the center of a sphere]

01:20

center of the sphere well that potential equals the amount of charge over 4 pi

01:25

times the electric constant times the radius since we're dealing with the

01:29

system after the spheres have reached equilibrium, their voltage will be the

01:32

same and so we can set the voltage equations for each sphere as equal also [Two voltage equations for each sphere]

01:37

we can cancel out that hole 4 pi electric constant jibber-jabber and just

01:42

find that charge one over radius 1 equals charge 2 over radius 2 which

01:48

means that the radius of the first sphere equals the radius of the second

01:51

sphere multiplied by the first charge over the second charge now because a little

01:55

something called the law of conservation of charge we know that the total charge [law of conservation of charge drawn on a table]

01:59

in the system has to remain the same the total charge equals the charge on the

02:04

first sphere before there was any transfer we can use that second equation Q sub 1

02:09

equals Q sub t minus Q sub 2 in the equation we just

02:13

saw for the radius and then we can plug in our numbers our sub 1 equals 12

02:18

centimeters times 8 coulombs minus 6 coulombs over six coulombs and that means

02:23

the radius of the first sphere equals four centimeters so the correct answer [equation for the radius of a sphere]

02:27

is B.. and remember if you're going to be an electrical entertainer well be careful

02:32

you don't want the people to be entertained by your pain [Sparky being electrocuted on stage]

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