Electric filed physics problem. Help

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Electrical field = 150 directed everywhere downward near Earth's surface.

1. What is the net electric charge on the Earth? Earth is a spherical conductor of radius 6371. km.
2. What is the electrostatic potential at the Earth’s surface, if the potential is taken to
be zero at infinity?
3. Using above info from Problem 1 Calculate the acceleration—magnitude and direction—of a proton released near the surface of the Earth. Disregard any interaction with air molecules.
4. Calculate the charge-to-mass ratio of a particle which would hover in place if released
near the surface of the Earth.

Part one and two I did can you check if its right?
1. q= Er^2 / k = -6.774*10^5 C
2. V= kq/r = -9.55*10^8 V

No idea about part three and four?


Thanks!
 
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i don't know if there is a force on it, but the electrical field of the earth, right?
wait i think I got it would I use:
F=ma the sub that in for E=F/q , giving a=Eq/mass of proton?
 
thanks! okay but I don't understand how to find charge to mass ratio?
 
okay I'm confused now there's something else besides the electric field of the earth?
 
is the Earth's mass involved
 
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sorry the protons mass multiplied by the 9.81
 
here's what I have so far but I'm not sure if its right:

I calculated acceleration due to the Earth's electric field a = Eq/m
so the force of this is F=mass of proton x acceleration I found above
then the other force is gravity? so F= mass of proton x 9.81
then add the two forces to find Fnet?
 
For the proton you should be able to show that the force of gravity is very much smaller than the electric force. So, you can forget about the force of gravity when calculating the acceleration of the proton.

But, for question 4, the force of gravity is important. How can the particle in question 4 hover at rest?
 
for it to hover at rest would the force be equal and opposite the electric field
 
Well, the two forces would have to be opposite to each other so that they cancel. This will tell you something about the sign of the charge of the particle.
 
i understand that the force of gravity and electric field will have to be opposite to each other, so the charged particle is negatively charged correct? then what about the mass of the particle how would i determine that.
is this equation relevant to use: ma=-Eq
 
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If the electric field is downward, then wouldn't the electric force also be downward if the charge is positive? But gravity also acts downward. So, there's no way the two forces could balance out to zero net force.

To determine the charge to mass ratio, write an equation that says that the magnitude of the force of gravity must equal the magnitude of the electric force.
 
I'm getting confused. So okay we have gravitational and electrical force both acting downward, so would the particle be negatively charged. and I am guessing this equation tells that "the magnitude of the force of gravity must equal the magnitude of the electric force": F=ma=Eq.
But now what is the value of the particle's charge would it be 1.602*10^-19 C ? and using the above equation if I don't know q how would i find m?
 
The gravitational force is of course downward. So, the electric force will need to be upward if the particle is to hover. Since E is downward, the sign of the charge will need to be negative.

You don't need to know the amount of charge. You are only asked to find the charge-to-mass ratio: q/m
 
F= ma and E=F/q So i sub, and get ma= Eq then solved for q/m
 
Good. You have the right equation. You just didn't manipulate it correctly to get the result for q/m.

Can you explain your steps in getting from ma = Eq to the result for q/m?
 
ma=Eq I divided by m on each side to get q/m= aE. Then a=-9.81 and E= -150. I multiplied. Oh wait okay so I was supposed to get: a/E= q/m. so It would be -9.81/-150 = -0.654. q/m = -0.654 Correct?
 
sept26bc said:
ma=Eq I divided by m on each side to get q/m= aE. Then a=-9.81 and E= -150. I multiplied. Oh wait okay so I was supposed to get: a/E= q/m. so It would be -9.81/-150 = -0.654. q/m = -0.654 Correct?

ok except for the location of the decimal point and you need to include proper units with your answer.
 
okay I fixed the units and decimal points on my paper. it would be -0.0654 C/kg .
okay if that is correct how would I find the total energy associated with the electric field of the Earth, if Earth is a charged capacitor? Would I use U=1/2 CV^2. Does the value of capacitance of the Earth matter?