Solve Point Charge Deflection: Potential Difference Between Plates

[waleye262]

Homework Statement

A point charge of mass 0.069 kg and charge q = +5.89 µC is suspended by a thread between the vertical parallel plates of a parallel-plate capacitor.

If the angle of deflection is 22°, and the separation between the plates is 0.025 m, what is the potential difference between the plates?

Homework Equations

U = .5C*V^2

The Attempt at a Solution

I know the left has a higher potential energy because the mass is attracted toward the right?

 

[Hootenanny]

Homework Statement

A point charge of mass 0.069 kg and charge q = +5.89 µC is suspended by a thread between the vertical parallel plates of a parallel-plate capacitor.

If the angle of deflection is 22°, and the separation between the plates is 0.025 m, what is the potential difference between the plates?

Homework Equations

U = .5C*V^2

The Attempt at a Solution

I know the left has a higher potential energy because the mass is attracted toward the right?

Welcome to PF,

If the charge is displaced to the right, then the left would indeed be at a higher potential.

Can you start by determining the tension in the string?

 

[waleye262]

The tension of the string is m*g= .68

I am still lost at how to find the the force that attracts it to the side.

would it be 5.89E-6 cos 22?

 

[Hootenanny]

The tension of the string is m*g= .68

That would be correct if the string was hanging vertically, but that isn't the case is it?

 

[waleye262]

no its not hanging vertically so mg=cos22?

I am sort of lost...

 

[Hootenanny]

no its not hanging vertically so mg=cos22?

You're getting closer:

T\cos22 = mg

Now since the mass is in equilibrium, what can you say about the horizontal forces acting on the mass?

 

[waleye262]

alright so the force acting on the mass is the same as the tension force which is 0.729.

so would you use F=q(v/x)?

 

[Hootenanny]

alright so the force acting on the mass is the same as the tension force which is 0.729.

so would you use F=q(v/x)?

Remember that you want the horizontal component of the force, not the vertical.