System involving gravity and electric field

In summary: 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.
  • #1
12
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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.
 
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  • #2
Lavas said:

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
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
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
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
The range between the two values worries, is that right?
 
  • #7
Lavas said:
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
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
Lavas said:
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].
 

What is the relationship between gravity and electric field?

The relationship between gravity and electric field is described by the fundamental forces of nature. Gravity is a force of attraction between objects with mass, while the electric field is a force of attraction or repulsion between charged particles. In some cases, these two forces can interact with each other, resulting in complex systems involving both gravity and electric fields.

How do gravity and electric field affect the motion of charged particles?

Gravity and electric fields can both affect the motion of charged particles. The strength and direction of the electric field can alter the path of a charged particle, while the gravitational force can cause the particle to accelerate towards a massive object. In systems involving both gravity and electric fields, the motion of charged particles can become quite complex.

What are some real-world examples of systems involving gravity and electric field?

One example of a real-world system involving gravity and electric field is the Earth's atmosphere. The electric field in the atmosphere can interact with gravity to cause lightning strikes. Another example is the motion of charged particles in a particle accelerator, where both gravity and electric fields are used to manipulate and accelerate particles.

How does the strength of the electric field affect the strength of the gravitational force?

The strength of the electric field does not directly affect the strength of the gravitational force. However, in some cases, the electric field can indirectly affect the gravitational force by altering the motion of charged particles, which can then influence the overall gravitational field in a system.

Can gravity and electric field cancel each other out?

In some cases, the forces of gravity and electric field can cancel each other out, resulting in a net force of zero. This can happen in systems where the strengths and directions of these two forces are equal and opposite. However, in most systems, the forces of gravity and electric field will interact with each other in complex ways and cannot be canceled out completely.

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