- #1

mayo2kett

- 23

- 0

so i used E=q/(Eo)A which gave me 150000 C/(8.89e-12 c^2/Nm^2)(.0200 m^2)=8.44e17 N/C

what do i do from here? how do i incorporate the 30° and and the weight of the ball??

-annie

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- Thread starter mayo2kett
- Start date

- #1

mayo2kett

- 23

- 0

so i used E=q/(Eo)A which gave me 150000 C/(8.89e-12 c^2/Nm^2)(.0200 m^2)=8.44e17 N/C

what do i do from here? how do i incorporate the 30° and and the weight of the ball??

-annie

- #2

LeonhardEuler

Gold Member

- 860

- 1

- #3

mayo2kett

- 23

- 0

so it would be like this? mass + k|q1||q2|/r^2\sin = E + mg/cos ???

which would give me... (6.70e-3)((((8.99e9)(150000)q2)/(r^2))/sin 30) = (8.436)(.0758)???

i would be solving for q2 right? but what about r^2, what would i make that?

-annie

which would give me... (6.70e-3)((((8.99e9)(150000)q2)/(r^2))/sin 30) = (8.436)(.0758)???

i would be solving for q2 right? but what about r^2, what would i make that?

-annie

Last edited:

- #4

LeonhardEuler

Gold Member

- 860

- 1

[tex] \vec{E}=\frac {q_{plate}} {A\epsilon_0} [/tex]

where [itex]q_{plate}[/tex] is the charge on the plates. The electric force is then

[tex]\vec{F}_{elec}=q_{ball}\vec{E}[/tex]

Equillibrium in the x direction means that the sum of the forces to the left is equal to the sum of the forces to the right:

[tex]\sum\vec{F}_{left}=\sum\vec{F}_{right}[/tex]

[tex]q_{ball}\vec{E}=\vec{T}cos(30)[/tex]

where[itex]\vec{T}[/tex] is the tension in the string. To find the tension force, you need to set up the equation for equillibrium in the y direction. In the y direction the sum of the up forces must equal the sum of the down forces. These will be the y component of the tension and the wieght, respectivly. From there just solve for [itex]q_{ball}[/tex] and your done.

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