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/sci/ - Science & Math

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5639042 No.5639042 [Reply] [Original]

I have a physics exam soon. What are some tricks or things to keep in mind when solving Gauss's Law and electric potential word problems? I always have trouble with those.

>> No.5639048

show tits

>> No.5639055

>>5639048
you suck

>> No.5639056

btw im a girl

>> No.5639059

give an example of a problem and what you have trouble with. Also tits.

>> No.5639065

My last exam has a conducting sphere made of layers of charge so the charge density changed with the radius of the sphere. It wanted to know the equation I would use to solve for the electric field inside the sphere.

>> No.5639067

And I'm not a girl...
>>5639056
this post wasn't me.

>> No.5639075

>>5639065
symmetrically charged Spherical shells have a rather unique and very useful property, which is that outside the shell you can treat it as a point charge. And inside the shell (and I mean anywhere inside, not just the center), the charge is 0. (due to all the forces canceling out and such just from the symmetry of the spherical shell).

As such, alot of gauss's law will have you make a spherical gaussian surface where you only consider charge from things inside of the spherical gaussian surface, while being able to let you disregard everything outside of it.

>> No.5639079

>>5639075
Would making it a non-conducting sphere change your answer to the electric field on the inside? I know the outside is always calculated as a point charge.

>> No.5639089

>>5639079
not at all, by the way you worded it I kind of knew you meant non conducting. You can think of any solid sphere as a infinitely many spherical shells.

Im guessing you had a problem like this "you have a sphere with a charge density of P, find the electric field as a function of r inside of the sphere"

What you do is at any given point r, you find the charge enclosed (in this case it is 4*pi*P*r^3/3). disregard all the charge outside of radius r, because you can just consider various spherical shells that you are inside of (hence their contribution to the field is 0). So Just treating the charge enclosed as a superimposed point charge, you get E=(charge enclosed)/(4*pi*episilon*r^2)=P*r/(3*episilon).

>> No.5639112

>>5639089
Oooh the way you put it makes sense to me.

I have another example question if you don't mind.

You have two equal charges q1 and q2. Charge q1is anchored to the floor while charge q2 is a block placed nearby. The block has a mass of m and a coefficient of friction x with the floor. When q2 is released how far will it slide before coming to a complete stop?

In this question I basically didn't know how to convert the work done by friction into a way to calculate distance traveled.

>> No.5639116

>>5639112
Work equals force * distance but the electrostatic force isn't constant.