Electric filed physics problem. Help!

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  • #26
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F= ma and E=F/q So i sub, and get ma= Eq then solved for q/m
 
  • #27
TSny
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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?
 
  • #28
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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?
 
  • #29
TSny
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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.
 
  • #30
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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?
 
  • #31
TSny
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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.
 
  • #32
vela
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Hint: If you want to find the acceleration, you use Newton's second law.
 
  • #33
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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
 
  • #34
vela
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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.
 
  • #35
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I don't understand the hovering part, would it be the gravitational force 9.81? what would I do then?
 
  • #36
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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?
 
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  • #37
SammyS
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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 [itex]\displaystyle \frac{|F_\text{electric}|}{|F_\text{gravity}|}=1\ .[/itex] Correct?

How is [itex]\displaystyle |F_\text{electric}| [/itex] related to E and [itex]\displaystyle |F_\text{gravity}| [/itex] related to g ?
 
  • #38
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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
 
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  • #39
SammyS
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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?
 
  • #40
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the acceleration would be zero because its not moving.
 
  • #41
vela
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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?
 

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