Electric flux through a sphere containing charges 8.0 and -4.9

[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.