Physics help...

[slevin_kelevra]

I have been working on this for a while and can't figure it out. Any help would be appreciated

What is the charge

inside the Gaussian surface?

Express your answer in terms of

(charge density),

, and any needed constants.

 

[slevin_kelevra]

it was supposed to say express in terms of lambda(charge density), and A(Area)

 

[OhMyBosh]

27

 

[4mica]

e = mc vagina

 

[spliff.Life]

were not that far yet, thats next month

 

[Chummer3]

so lambda= q/A

the cross sectional area of the plane is pi*r^2

solve for q, try q=lambda*(pi r^2)

 

[moonofendor]

so the surgace area intergal equals q over epsilon

$Eda= q/e

E is constant so pull that outta the intergal

so

E$da=q/e

da will be equal to the area

so

Epir^2=q/e

q=L/e

L=q/a

a=pir^2

so q=Lpir^2

so

Epir^2=Lpir^2/e

the pir^2 cancel out

so

E=L/e

or

E=q/pir^2e

or

E=2kq/r^2

 

[spruitm]

GREEN yeaaa

 

[Scott~]

Pretty sure this is incorrect because the two sides dont cancel out, net flux is positive on both ends so it will actually be twice the area. Also you solved charge as though this was a thin line of charge instead of being an infinite sheet. Which is sigma(charge density) * A.

2 pi r^2 * E = sigma*A/epsilon

You can simplify and manipulate this if youd like, but since you werent actually given anything there isnt a lot you can do.

 

[NESblowjob]

i think its just charge density over the area of the top of the cylinder

 

[An10a_Eng]

you = mc fucking faggot

 

[Hansel.]

haha thank you for making a funny

 

[slevin_kelevra]

thanks +k

the answer was q=(charge density)*(Area)