Equilibrium Position of Charged Ball near Charged Plate

In summary: Hello VictorWutang,In summary, a small positively charged sphere of mass m is attached to a very large positively charged vertical plate by a string and hangs due to gravity. The charge density on the plate, σ, is uniform. The equilibrium position of the ball as a function of the (uniformly distributed) charge on the ball is found using the Pythagorean theorem. The x component of tension is sinθ * Ftension and the y component is cosθ * Ftension, where Ftension is the tension in the string and θ is the angle between the wall and the rope. The electrical field is E = .5kσ/r and the electrical force is F = qE. The capacitance between the sphere and
  • #1
VictorWutang
13
0

Homework Statement



A small positively charged sphere of mass m is attached to a very large positively charged vertical plate by a string and hangs due to gravity. The charge density on the plate, σ, is uniform.

a) Calculate the equilibrium position of the ball as a function of the (uniformly distributed) charge on the ball. I.e. what angle does the string make with the vertical?

b) estimate the capacitance between the sphere and the plate.

Homework Equations



F = kqσ / r2 (not sure if this is correct)
E = kq / r

The Attempt at a Solution



a) Pythagorean theorem will come into use. I have to find where the force due to the magnetic field of the plate is equal to the force of gravity resisting it, but I can't figure out how to make the force of gravity,the force of the electric field and the tension/normal force to be in vectors that cancel out. More specifically, i can't figure out how to express the downward force of gravity to be perpendicular to the plate to cancel out the force from the plate.

b) once i find this distance, capacitance will just be ε0*A / d with A being ∏r2 with r being the radius I would estimate the ball to be.
 
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  • #2
VictorWutang said:

Homework Statement



A small positively charged sphere of mass m is attached to a very large positively charged vertical plate by a string and hangs due to gravity. The charge density on the plate, σ, is uniform.

a) Calculate the equilibrium position of the ball as a function of the (uniformly distributed) charge on the ball. I.e. what angle does the string make with the vertical?

b) estimate the capacitance between the sphere and the plate.


Homework Equations



F = kqσ / r2 (not sure if this is correct)
E = kq / r


The Attempt at a Solution



a) Pythagorean theorem will come into use. I have to find where the force due to the magnetic field of the plate is equal to the force of gravity resisting it, but I can't figure out how to make the force of gravity,the force of the electric field and the tension/normal force to be in vectors that cancel out. More specifically, i can't figure out how to express the downward force of gravity to be perpendicular to the plate to cancel out the force from the plate.

b) once i find this distance, capacitance will just be ε0*A / d with A being ∏r2 with r being the radius I would estimate the ball to be.

Hello VictorWutang,
For the first part the tension acts as the balancing force.Its horizontal component balances the electric force due to plate and the vertical component balances gravity.
regards
Yukoel
 
  • #3
So is the x component of tension is sinθ * Ftension?

and then Ftension is mass * gravity?
 
  • #4
VictorWutang said:
So is the x component of tension is sinθ * Ftension?

and then Ftension is mass * gravity?

Hello VictorWutang,
The tension needs to be resolved both vertically and horizontally right?
So Ftension is mass * gravity?[/QUOTE]
doesn't sound correct.Would you put a picture of where you take your angle from?
|
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| \
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I think the oblique slashes represent the string in your diagram the vertical slashes the plate and 8 the charge .
Please check again.
EDIT: Spaces were automatically truncated in my reply.So apologies for the confusing diagram .
regards
Yukoel
 
  • #5
yes, that's what the diagram looks like. and θ is the angle I'm looking for, the angle between the wall and the rope
 
  • #6
VictorWutang said:
yes, that's what the diagram looks like. and θ is the angle I'm looking for, the angle between the wall and the rope
Hello again,
Then wouldn't Tsinθ=Electrical force(Calculate)
and Tcosθ=Weight ?
You can eliminate tension from your equations wouldn't you?


regards
Yukoel
 
  • #7
so electrical field is E = .5kσ/r and electrical force is F = qE so

Tsinθ = F = qE = .5qkσ/r, also Tcosθ= mg giving us T = mg/cosθ so

mgtanθ = .5qkσ/r so

θ = tan-1 (.5qkσ/rmg) but r in this case is the distance not the radius, and we want to put θ in terms of q being variable

so how can i put r in terms of q?-----also, if you could help with the other two problems i posted i'd really appreciate it. you're a big help on this one too.
 
  • #8
VictorWutang said:
so electrical field is E = .5kσ/r and electrical force is F = qE so

Tsinθ = F = qE = .5qkσ/r, also Tcosθ= mg giving us T = mg/cosθ so

mgtanθ = .5qkσ/r so

θ = tan-1 (.5qkσ/rmg) but r in this case is the distance not the radius, and we want to put θ in terms of q being variable

so how can i put r in terms of q?


-----also, if you could help with the other two problems i posted i'd really appreciate it. you're a big help on this one too.
Hello again,
The field due to an infinitely long sheet of charge with a given charge density σ
is given by (σ/(2ε_o))It is not a function of r.Your expression for E when revised would lend you the desired result.
regards
Yukoel
 
  • #9
Yukoel said:
Hello VictorWutang,
Would you put a picture of where you take your angle from?
|
| \
| \
| \
| \
| 8
I think the oblique slashes represent the string in your diagram the vertical slashes the plate and 8 the charge .
Please check again.
EDIT: Spaces were automatically truncated in my reply.So apologies for the confusing diagram .

Yukoel

I draw a picture with Paint and uploaded. It looks nicer than that made of characters :smile:

ehild
 

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FAQ: Equilibrium Position of Charged Ball near Charged Plate

What is the equilibrium position of a charged ball near a charged plate?

The equilibrium position of a charged ball near a charged plate is the point at which the forces acting on the ball are balanced, causing it to remain stationary. This occurs when the electrostatic force between the ball and the plate is equal and opposite to the force of gravity.

How is the equilibrium position affected by the magnitude of charge on the ball and the plate?

The equilibrium position is directly proportional to the magnitude of the charges on both the ball and the plate. This means that as the charges increase, the equilibrium position shifts further away from the plate.

What happens to the equilibrium position if the distance between the ball and the plate changes?

The equilibrium position is inversely proportional to the distance between the ball and the plate. This means that as the distance increases, the equilibrium position moves closer to the plate, and vice versa.

Can the equilibrium position be affected by the presence of other charged objects or external forces?

Yes, the equilibrium position can be affected by the presence of other charged objects or external forces. These additional forces can cause the equilibrium position to shift or the ball to move away from the plate.

How can the equilibrium position be calculated?

The equilibrium position can be calculated using Coulomb's law, which states that the electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. By setting the electrostatic force equal to the force of gravity, the distance at which they are balanced can be determined as the equilibrium position.

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