8th October 2011 - 01:43 PM
I have to take this physics class and there I have to solve this problem using Gauss' law.
The earth (here seen as a conducting ball) has got an electric charge. The
resulting field can be detected at the surface with a sensitive electric
measurement device and has got average value of 150 N/C and points in
direction of the geocenter.
NOTE: Do not use the radius of the Earth.
a) What is the sign of the charge of the earth?
b.) How big is the surface charge density of the earth?
c) How much is the overall charge of the earth?
d) what would change, if the earth would be assumed as perfectly insulating ball,
but with the same overall charge?
e) Calculate the earth's electric field at the surface for this case
how I tried:
a) since the direction is towards the center, Earth is negatively charged.
b.) Surface charge density I can get by multiplying epsilon_0 and the given field (-150N/C)
c) This I would have solved by multiplying the surface charge density and 4.pi.r^2
But its mentioned I shouldn't use the Radius. Can I somehow find the radius from the problem itself?
d) Nothing would change I think, only that the charge wont be on the surface but distributed through the ball.
e) This is my main problem. Lets say I have calculated the charge in sec c), how would I again calculate the field without the radius. My friend suggested I can do something with integrating volume charge density, but I always end up with an R.
Is there a way like that?
8th October 2011 - 04:42 PM
The game here is to use Gauss' Theorem. In the case where the charge distribution is spherically symmetric and wholly contained within a sphere of finite radius, the electric field outside that radius depends only on the total charge. Whether that charge is distributed over a spherical shell, a spherical volume or concentrated at a point, as long as the symmetry is maintained the electric field outside the radius is unchanged.
This should give you c and e trivially.
8th October 2011 - 08:26 PM
Thanks very much for the prompt reply. I understand that I have to use Gauss theorem but somehow I am not able to deduce the Electric field of an insulating Earth without the use of the Radius of the Earth, for question e). All the formula I get is E = q/4.pi. epsilon_0. r^2. I am not able to think like a physicist as I am totally not used to this kind of things.
I understand that we have a finite radius and symmetric charge distribution, but how will the radius cancel out in c and e?
please let me know,
thanks and regards,
9th October 2011 - 09:18 PM
QUOTE (Kino+Oct 8 2011, 04:42 PM)
the electric field outside that radius depends only on the total charge.
...is the key point. They gave you the E field with one charge distribution; they didn't change the total charge, so the E field is the same. Part e, in no easy steps - the beauty of Gauss' Theorem.
I'm actually not sure how to get the total charge without using the radius, however, despite my earlier certainty. If you've done it, I'd be interested to see your answer. If not, I'd be interested to see your lecturer's answer.
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