Potential difference of plates

In summary, the sphere hangs by a thread between two parallel plates with uniform surface charge densities (+ and -). The charge on the sphere is 8.90 10-6 C. If the potential difference between the plates is 47.7 V, the sphere will assume an angle of 30.0° with the vertical.
  • #1
ovoleg
94
0
Can anyone help me with a hint of some sort? I am lost...

A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 cm apart. The plates are insulating and have uniform surface charge densities + and -. The charge on the sphere is q = 8.90 10-6 C. What potential difference between the plates will cause the thread to assume an angle of 30.0° with the vertical

24_37.gif
 
Last edited:
Physics news on Phys.org
  • #2
I know the field between two charged plates is E=sigma/epsilon_0...

The way I think of it is that I need to find the forces acting on it and then have that be my E value and then integrate E through distance 5cm?

This is what I tried but I did not get 47.7V

kq*cos30*.015kg*.05=51.95V
 
  • #3
a) What is the definition of the electric field in terms of voltage (potential difference) and distance ? There is a simple formula, check your text or notes.

b) What is the relationship between the force on a charged body, the quantity of charge on that body and the electric field strength the body is subjected to ?

c) Draw a force diagram. What are the forces acting on the ball, and how do they resolve horizontally and vertically ?

d) Put it all together (in symbols), rearrange to form an expression for V, the voltage, then plug in values to get the answer.
 
  • #4
Curious3141 said:
a) What is the definition of the electric field in terms of voltage (potential difference) and distance ? There is a simple formula, check your text or notes.

b) What is the relationship between the force on a charged body, the quantity of charge on that body and the electric field strength the body is subjected to ?

c) Draw a force diagram. What are the forces acting on the ball, and how do they resolve horizontally and vertically ?

d) Put it all together (in symbols), rearrange to form an expression for V, the voltage, then plug in values to get the answer.

a) E=Vab/d
b) E = F/q

c)F=(mg)*cos(30)?

d) Vab=(mg)*cos(30)*d/q?

I get 71.52007V

:(

Please help :) I've been trying for the past few hours
 
Last edited:
  • #5
Should be tan(30) right? Draw the force diagram again. Other than that, I think you've got it.
 
  • #6
durt said:
Should be tan(30) right? Draw the force diagram again. Other than that, I think you've got it.
'
THANKS!
 
  • #7
ovoleg said:
a) E=Vab/d
b) E = F/q

Those two are fine. :smile:


c)F=(mg)*cos(30)?

No. OK, the sphere is in equilibrium right ? So there is no net force on i, all the forces balance. What are these forces ?

Let's list them :

1) Weight = mg, acts vertically downward.
2) Tension in the suspending string, call it T, acts upward in the direction of the string (at the 30 degree angle it is hanging at).
3) Electrostatic force, call it F = qE = qV/d, where q is the charge and E the electric field strength, V is the voltage and d is the distance between the plates. Acts horizontally to the right.

The easiest thing to do now is to resolve the forces horizontally and vertically and set up equations.

Basic concept : Vertical component of tension exactly balances the weight.
Horizontal component of tension exactly balances the electrostatic force.

You can get two equations with two unknowns, T and V. You need to solve for V so eliminate T. Now get an expression for V.

d) Vab=(mg)*cos(30)*d/q?

Not correct, I'm afraid. Hopefully, you can get it now. Don't forget to specify which plate is positively charged, left or right. Remember the ball is positively charged, according to the question.
 
  • #8
durt said:
Should be tan(30) right? Draw the force diagram again. Other than that, I think you've got it.

Hold on, I think it's important he gets this conceptually instead of just trying out different trig ratios.
 

What is the potential difference of plates?

The potential difference of plates, also known as voltage, is the difference in the electric potential energy per unit charge between two points in an electric field. It is measured in volts (V) and is a measure of the intensity of the electric field.

How is the potential difference of plates calculated?

The potential difference of plates can be calculated by dividing the work done (in Joules) in moving a unit positive charge from one plate to another by the amount of charge (in Coulombs). This can be represented by the equation V = W/Q, where V is the potential difference, W is the work done, and Q is the amount of charge moved.

What factors affect the potential difference of plates?

The potential difference of plates is affected by the distance between the plates, the magnitude of the charges on the plates, and the dielectric material between the plates. It is also affected by external factors such as the presence of other charges or external electric fields.

What is the relationship between potential difference and electric current?

The potential difference between two points is directly proportional to the electric current flowing between them, according to Ohm's law. This means that an increase in potential difference will result in an increase in current, and vice versa, as long as the resistance remains constant.

Why is potential difference important?

Potential difference is important because it is a key factor in determining the flow of electric current in a circuit. It is also crucial in understanding the behavior of charged particles in an electric field and plays a significant role in various applications such as electricity generation and distribution, electronics, and electrochemistry.

Similar threads

Replies
1
Views
144
  • Introductory Physics Homework Help
Replies
14
Views
829
  • Introductory Physics Homework Help
Replies
6
Views
314
  • Introductory Physics Homework Help
Replies
11
Views
392
  • Introductory Physics Homework Help
Replies
26
Views
576
  • Introductory Physics Homework Help
2
Replies
58
Views
3K
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
730
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Back
Top