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System involving gravity and electric field

  • Thread starter Lavas
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  • #1
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Homework Statement


A ball of mass .01 hangs 1.5m with a charge of 3u(micro)C from a thin string attached to the top of a wall with uniform charge density .3uC. At what angle of the string from the wall at a stable equilibrium


Homework Equations


Gauss' Law Integral (E * A) = Q/\[Epsilon]


The Attempt at a Solution


Went somewhere with it but not really sure and already turned it in, too tired to try again.
 

Answers and Replies

  • #2
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Homework Statement


A ball of mass .01 hangs 1.5m with a charge of 3u(micro)C from a thin string attached to the top of a wall with uniform charge density .3uC. At what angle of the string from the wall at a stable equilibrium


Homework Equations


Gauss' Law Integral (E * A) = Q/\[Epsilon]


The Attempt at a Solution


Went somewhere with it but not really sure and already turned it in, too tired to try again.
Hi Lavas,

What are the units of the "0.01" mass? What about the charge density of the wall. Is the total charge 0.3 uC, or is it 0.3 uC/m2 (or something else)?

Also, please show where you went with it so far! :smile:
 
  • #3
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Sorry it is .01kg and 0.3 uC/m2
Well I think at least the electric field coming off the wall = Sigma/[2(Epsilon)]
The charge from the from the mass = q/[(4)(Pi)(r^2)(Epsilon)]
mg= 1N
but honestly from there is was just guesswork and I can't remember what I put.

(I'm doing this problem from memory on an assignment I've already turned it and just wanted to see what the answer actually was)
 
  • #4
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I can't give you the answer, but perhaps I can get you started. So, if the electric field coming off the wall is

[tex] E = \frac{\sigma}{2 \epsilon_0} [/tex]

then the horizontal, electrical force on the ball is

[tex] F = qE = q \frac{ \sigma}{2 \epsilon_0} [/tex]

where q is the charge on the ball.

Of course, since everything is in equilibrium (i.e. not accelerating), this horizontal, electrical force must be counteracted by the horizontal component of the tension on the string. Then you can (and you must) also bring in other tension (components) and gravitational forces, after which you can solve for the angle. :wink:
 
  • #5
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So that makes q \frac{ \sigma}{2 \epsilon_0} = 5.0824486*(10*10)
and mg =.01*9.8=.098
so would the tension equal the vector sum of the mg and electric field
 
  • #6
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The range between the two values worries, is that right?
 
  • #7
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So that makes q \frac{ \sigma}{2 \epsilon_0} = 5.0824486*(10*10)
and mg =.01*9.8=.098
so would the tension equal the vector sum of the mg and electric field
Something is not quite right with your electric force.

[tex] q = 3 \times 10^{-6}\ C [/tex]

[tex] \sigma = 0.3 \times 10^{-6} \ \frac{C}{m^2} [/tex]

[tex] \epsilon_0 = 8.854 \ \times 10^{-12} \ \frac{C^2}{Nm^2} [/tex]

Try calculating the force

[tex]
F = q \frac{ \sigma}{2 \epsilon_0}
[/tex]

with these values.
 
  • #8
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Long day, stupid mistake but that gave me .10164N but I'm still confused as whether the tension is the vector sum of Mg and Horizontal electric force. Also how that applies to the angle.
 
  • #9
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Long day, stupid mistake but that gave me .10164N but I'm still confused as whether the tension is the vector sum of Mg and Horizontal electric force. Also how that applies to the angle.
You forgot to divide by the 2. :tongue: The electric force should be 0.05082 N.

But yes, the tension on the string is essentially the vector sum of the vertical gravitational force and the horizontal electric force, but in the opposite direction (at least from the point of view of the ball). Since nothing is accelerating (i.e. everything is in equilibrium), the vertical component of the tension = mg, but points up instead of down. The horizontal component of the tension equals the electric force, and points toward the wall.

Use sines and cosines of [tex] \theta [/tex] to break up the tension, T, into its components. Then you'll have a total of 2 unknowns (T and [tex] \theta [/tex]). And fortunately, you'll have two equations, so you can solve for [tex] \theta [/tex].
 

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