Electric filed physics problem. Help

[sept26bc]

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!

 

[TSny]

Hi sept26bc. In general, what causes a particle to accelerate?

 

[sept26bc]

a force

 

[TSny]

Is there a force on the proton? How would you calculate that force?

 

[sept26bc]

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?

 

[TSny]

Yes!

 

[sept26bc]

thanks! okay but I don't understand how to find charge to mass ratio?

 

[TSny]

Well, there's actually another force acting. And what really matters is the net force.

 

[sept26bc]

force of gravity?

 

[TSny]

Sure.

 

[sept26bc]

okay I'm confused now there's something else besides the electric field of the earth?

 

[TSny]

Yes. Earth also has a gravitational field. How do you calculate the force due to gravity?

 

[sept26bc]

is the Earth's mass involved

 

[TSny]

No! Not the Earth's mass.

 

[sept26bc]

sorry the protons mass multiplied by the 9.81

 

[TSny]

Good. Try calculating it and compare to the electric force.

 

[sept26bc]

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?

 

[TSny]

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?

 

[sept26bc]

for it to hover at rest would the force be equal and opposite the electric field

 

[TSny]

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.

 

[sept26bc]

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

 

[TSny]

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.

 

[sept26bc]

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?

 

[TSny]

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

 

[TSny]

That's not quite right. Can you write the equation you started with to get this?

 

[sept26bc]

F= ma and E=F/q So i sub, and get ma= Eq then solved for q/m

 

[TSny]

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?

 

[sept26bc]

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?

 

[TSny]

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.

 

[sept26bc]

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?

 

[TSny]

Hint: Write U as U = 1/2 (CV) V and recall that CV = Q for a capacitor. So, you can express U in terms of Q and V (which you have already calculated) and you won't need a value for C.

Or, you could use Q = CV to find C and then plug into your equation.

 

[vela]

Hint: If you want to find the acceleration, you use Newton's second law.

 

[sept26bc]

Okay Thanks! I found acceleration can you reassure my answer?
so F=ma and E=F/q I substitute and get a= Eq/m and with the numbers plugged in I get:
a= (-150N/C)(1.602*10^-19 Nm^2/C^2) / (1.67*10^-27 kg)
a= -1.44*10^10 m/s^2

for charge to mass ratio I used E=ma/q rearrange to get q/m= a/E. Plug in the numbers of a and E, (-1.44*10^10m/s^2)/ (-150N/C) = q/m answer is -0.0654 kq/C

Another question is how would I find the total energy associated with electrical field of the Earth if the Earth is like a charged capacitor? Would I use U=1/2CV^2 ?

thanks

 

[vela]

You didn't do part 4 correctly. What are the forces on the particle and what is its acceleration if it's hovering?

If you're considering the Earth a capacitor, then yes, you can use that formula. You'll need to figure out the capacitance of the Earth.

 

[sept26bc]

I don't understand the hovering part, would it be the gravitational force 9.81? what would I do then?

 

[sept26bc]

Vela. I'm really confused I've been working on this for hours. I know that the gravitational and electric force have to cancel in order for the particle to be hovering.how would another force be included in the equation q/m=a/E ? or am i using the wrong equation?

 

[SammyS]

Vela. I'm really confused I've been working on this for hours. I know that the gravitational and electric force have to cancel in order for the particle to be hovering.how would another force be included in the equation q/m=a/E ? or am i using the wrong equation?

So, you need \displaystyle \frac{|F_\text{electric}|}{|F_\text{gravity}|}=1\ . Correct?

How is \displaystyle |F_\text{electric}| related to E and \displaystyle |F_\text{gravity}| related to g ?

 

[sept26bc]

well F=ma and E=F/q so E= ma/q

|Felectric| is related to E by E=F/q
and |Fgravity| related to g by F=ma

correct?

im not understanding why you have Felectric/ Fgravity=1

 

[SammyS]

well F=ma and E=F/q so E= ma/q

|Felectric| is related to E by E=F/q
and |Fgravity| related to g by F=ma

correct?

im not understanding why you have Felectric/ Fgravity=1

If the particle hovers, what is its acceleration?

 

[sept26bc]

the acceleration would be zero because its not moving.

 

[vela]

So you know there are two forces on the particle, the force of gravity $\vec{F}_g$ and the electric force $\vec{F}_E$. You also know the acceleration is 0 because it's not moving.

What is the magnitude of $\vec{F}_g$? What is the magnitude of $\vec{F}_E$?

Now take all that info and use Newton's second law, which is $\sum \vec{F} = ma$. What do you get?