Opposite Charged Plates with Sphere

In summary, to solve for the potential difference between two parallel plates causing a thread to assume an angle of 30 degrees with the vertical, the following equation can be used: Delta(V) = Ed = (mgd*tan(30))/q, where m is the mass of the hanging sphere converted to kg, g is the acceleration due to gravity, d is the distance between the plates, and q is the charge on the sphere. The charge should be in Coulombs and the distance in meters. Also, to convert from degrees to radians, multiply degrees by pi/180.
  • #1
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Homework Statement


A small sphere with mass 2.90 g hangs by a thread between two large parallel vertical plates 0.05 m apart. The plates are insulating and have uniform surface charge densities + \sigma and - \sigma . The charge on the sphere is q = 9.70×10^6 C.

What potential difference between the plates will cause the thread to assume an angle of 30.0 deg with the vertical

Homework Equations



[tex] Delta(V) = Ed = \frac {F_{elec} * d} {q} = \frac {(mg)*(d)*tan(30)} {q} [/tex]

The Attempt at a Solution



[tex] \frac {(2.90*8.90)*(0.05)*tan(30)} {9.70*10^{-6}} [/tex]

The online program complains that its wrong, I'm wondering is the equation correct?
 
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  • #2
stylez03 said:

Homework Statement


A small sphere with mass 2.90 g hangs by a thread between two large parallel vertical plates 0.05 m apart. The plates are insulating and have uniform surface charge densities + \sigma and - \sigma . The charge on the sphere is q = 9.70×10^6 C.

What potential difference between the plates will cause the thread to assume an angle of 30.0 deg with the vertical

Homework Equations



[tex] Delta(V) = Ed = \frac {F_{elec} * d} {q} = \frac {(mg)*(d)*tan(30)} {q} [/tex]

The Attempt at a Solution



[tex] \frac {(2.90*8.90)*(0.05)*tan(30)} {9.70*10^{-6}} [/tex]

The online program complains that its wrong, I'm wondering is the equation correct?

g=9.8 not 8.9?
 
  • #3
lylos said:
g=9.8 not 8.9?

That was actually a typo, I used 9.8 and still no luck.

The computation comes out to:

[tex] -9.38*10^{-7} [/tex]

Can anyone confirm this?
 
Last edited:
  • #4
Convert 2.90 from g to kg. Also you are using 9.70×10^-6 C in the denominator, but the initial value was 9.70×10^6 C.
 
  • #5
ranger said:
Convert 2.90 from g to kg. Also you are using 9.70×10^-6 C in the denominator, but the initial value was 9.70×10^6 C.

The mass was what threw it off, actually Q was 10^-6, I guess copying it over from the flash page removed the negative, though I had that in my solution. Also I figured out the answer. THanks!
 
Last edited:
  • #6
http://www.krellinst.org/UCES/archive/resources/trig/node10.html" [Broken]

To convert from degrees to radians, multiply degrees by pi/180.

To convert from radians to degrees, multiply by 180/pi.
 
Last edited by a moderator:

1. What is the purpose of using opposite charged plates with a sphere?

The purpose of using opposite charged plates with a sphere is to create an electric field between the plates, which can be used for various scientific experiments and demonstrations.

2. What is the effect of placing a sphere between two opposite charged plates?

When a sphere is placed between two opposite charged plates, it will experience an electric force due to the difference in charge between the plates. This force will cause the sphere to move towards the plate with the opposite charge.

3. How does the distance between the plates affect the electric field strength?

The distance between the plates has a direct effect on the strength of the electric field between them. The closer the plates are, the stronger the electric field will be. This is because the electric field lines are closer together, resulting in a higher electric field strength.

4. Can the electric field between opposite charged plates be uniform?

Yes, it is possible to create a uniform electric field between opposite charged plates. This can be achieved by using plates with a large surface area and keeping the distance between them constant.

5. What happens to the electric field if the charges on the plates are changed?

If the charges on the plates are changed, the electric field strength between them will also change. Increasing the charge on the plates will result in a stronger electric field, while decreasing the charge will result in a weaker electric field.

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