Solve Gauss's Law: 8.0 & -4.9 Charges in Uncharged Sphere

[bmandude156]

Homework Statement

A 8.0 charge and a -4.9 charge are inside an uncharged sphere.
What is the electric flux through the sphere?

Homework Equations

It says in the book that Electric flux = E(4piR^2) although I do not know the radius and yet still don't believe that's the way to go to solving this problem.

The Attempt at a Solution

I attempted to add the charges together because I had no other idea what to do with them. And then tried to use the equation of a sphere for electric flux which is above. I do not know for sure if that is the way to proceed but I did not get the answer correct.

 

[sgd37]

you don't need the radius that's the point of gauss's law. Just substitute coulombs law for the electric field i.e. \frac{q}{4 \pi \epsilon_0 r^2} into your equation for flux

 

[bmandude156]

I have immediate reservations after seeing your reply for i still see r^2 in your denominator. Isnt that going to be a problem and as for the charges, should I merely take the sum of them to be my q.

 

[betel]

sgd is talking about the electric field. You have to do something to it to get the electric flux.

 

[bmandude156]

i believe the equation at the moment is Electic flux = q/4piEnotr^2. by doing something I think ur implying..doing something to get rid of r^2. but i haven't a clue how or if that's what ur suggesting i do. sgd said that i didnt have to worry about r anymore but that's not true. coloumbs law had one itself. I am confused

 

[betel]

What are the quantities entering the formula for the electric flux. Your last formula is not correct.

 

[bmandude156]

well now I am more confused than ever. sgd gave me the formula. so r u saying that his formula was wrong.

 

[kaksmet]

In your first formula, you have the electric field, E, on the right hand side. See if you can use sgd37 formula for the electric field to calculate your flux.

 

[betel]

Do you know the difference between electric field and electric flux?

 

[bmandude156]

electric flux=E(4piR^2) and Electic flux = q/4piEnotr^2(was sgd's equation). now, what i think ur saying is take sgd's equation of electic flux and put it into mine. so the result is as follows: E(4piR^2) = q/4piEnotr^2. what i hoped would happen is by doing this, is alregbriaclly get rid of R. but i don't think that could happen. and for that matter...what is E now?!?...im afraid this is diogn more harm than good

 

[betel]

Perhaps you should first think about the equations and not randomly plugging one into the other that has the same letters in it.

 

[bmandude156]

you don't need the radius that's the point of gauss's law. Just substitute coulombs law for the electric field i.e. \frac{q}{4 \pi \epsilon_0 r^2} into your equation for flux

. I am must be misunderstanding what the equation looks like than

 

[betel]

I have to repeat my question.

Do you know the difference between electric field and electric flux?

We can continuer argueing forever but if you don't know the difference we will get nowhere.

 

[bmandude156]

Electric flux is proportional to the number of electric field lines going through a surface. NO arguing here mate. just need help. electric field is like the magnetic field that pertains to charges attraction with each other

 

[betel]

Well, that is good.
Now what does Gauss's Law state? You already refer to it in the title, so why don't you do it the easy way?

 

[bmandude156]

Electric flux = E(4piR^2)..that one? But there was a problem with it remember...r^2?

 

[betel]

Now what is E in that formula?

Maybe Gauss's law in words is easier here.

 

[bmandude156]

electric field...sorry it took so long for me to respond...didnt see ur response on tihs page

 

[betel]

And the formula for the electric field is?

 

[bmandude156]

E=F/q

 

[betel]

I'm not sure what your F and q are but this formula seems strange.

 

[bmandude156]

hahah right from the book in my previous chapter. what is urrr formula for the electric field so we can progress?

 

[betel]

Then tell me what F and q are supposed to mean. I dont't know every definition of every book by heart and without that letters are meaningless.

And btw. I dont' have to solve this question.

 

[bmandude156]

F is the electric force and q is as usual, the charge. perhaps this isn't the proper equation seeing as ur unfamiliar with it

 

[betel]

Well it is not exactly the equation that will be of much help to us.

Didn't you derive a formula for the electric field of a point charge? I.e. if you have a charge q at radius r, what is the strength of the electric field?

And I still think you should have a statement of Gauss's law in words. This would kill this question in one go.

 

[bmandude156]

I will try to reply to ur post ASAP and i appreciate ur post. But ironically, I have a physics class at the moment. brb later

 

[sgd37]

wow this is one funny thread

so for the sake of my sanity; denoting electric FLUX with the Letter \Phi and the electric FIELD as E

Now \Phi = E 4 \pi r^2 and E = \frac{q}{4 \pi \epsilon_0 r^2}

Substituting for E in \Phi you have

\Phi = \frac{q 4 \pi r^2}{4 \pi \epsilon_0 r^2} = \frac{q}{ \epsilon_0}

This is Gauss's law that the flux through a surface is the charge encompassed by that surface divided by the vacuum permittivity

 

[bmandude156]

ALRIGHT SO... what about the two charges. should i just sum them

 

[sgd37]

yeah

 

[bmandude156]

...but that didnt work good sir. the answer was incorrect. i got 3.5 X 10^-11.

 

[sgd37]

I'm pretty sure Gauss is correct. But what units are you dealing with. If the charges are in coulombs then you should have a flux of 3.5 * 10^11 Vm which seems ridiculous. But if they were in terms of the electron charge then the flux would be 5.6 * 10^-8 Vm

 

[bmandude156]

KN*m2/C is the units...i tried 3.5 x 10^-6 but it failed

 

[sgd37]

i meant the units for charge. What is the answer supposed to be

 

[bmandude156]

the unit i mentioned. the KN*m^2/C

 

[bmandude156]

the charges are in mirco coloumbs

 

[betel]

Seems I missed the continuation of this thread.

Just as a note. Adding the charges and acting as if they are just one charge at the origin works in this case because of gauss law for the flux. The total electric field is not simply the formula sgd gave with the total charge but would be the sum of two terms each representing one of the point charges. Then integrating over a sphere would not be easy.

Which is why I would have used the worded form of Gauss's law.

The electric flux through any closed surface is equal to the total charge in the volume inside this surface.

For the units...

 

[betel]

You can never get the unit you gave. The Vm sgd gave is the correct unit.

 

[bmandude156]

but how sir when i got that answer wrong lol. those units are what my teacher expects the answer in.

 

[betel]

Does the numerical value of the answer fit? Look up SI units. There is no way to convert
kiloNewton* meter^2 / Coloumb to
Volt * meter

 

[sgd37]

they're equivalent, apart from the kilo bit of course but that isn't a problem, both denote flux

 

[bmandude156]

i believe u betel when u say it cannot be converted into those units. but surely my teacher would not have been so unreasonable as to do that to us. surely there is a way around this

 

[betel]

sgd is right.
Seems i wasn't really awake when I tried to convert the units. But converting the units will only change factors of 10.

 

[bmandude156]

so is my answer still 3.5 X 10^-11 or something else

 

[betel]

No it is 3.5\cdot 10^{11} Vm = 3.5\cdot 10^8 KN m^2/C.
And from the data you gave (charges 8 mC and -4.9 mC) this is correct. I agree with sgd on that.

 

[bmandude156]

i think u mean 3.5 * 10^-8 sir

 

[betel]

\Phi=\frac{q}{\epsilon_0}=\frac{3.1\mu C}{8.85418782 \cdot 10^{-12} \frac{As}{Vm}}=0.35 \cdot 10^6 Vm=3.5\cdot 10^5 Vm = 3.5\cdot 10^2 kN m^2/C

I had calculated with milli Coulomb before.

 

[bmandude156]

fair enough...anddddddddddddddddddddddddd yayaaaa that's the answer...thankyou betel and sgd. you guys are champs. i appreciate your help.