What is the weight of two suspended conducting spheres with given charges?

[Kudo Shinichi]

HELP! A electricity problem

Homework Statement

Two small identical conducting spheres are each suspended by a 0.50m long light thread from a common point. when charges of +2.0$$\mu$$C and +3.0$$\mu$$C are placed on the spheres they are in equilibrium when separated by 0.60m. Find the weight of the spheres.

The Attempt at a Solution

I started with Coulomb's law
F=(k*q1*q2)/r^2
k=8.99*10^9
q1=+2.0$$\mu$$C=2*10^-6C
q2=+3.0$$\mu$$C=3*10^-6C
r=0.6
therefore, F=0.15N
The question asks me to find the mass for both spheres,since they are two small identical conducting sphere, they should have same mass
so, F=ma,0.15=ma or F=mg, 0.15=9.8m
However, I don't really think it is the right way to solve this problem

Thank you for helping me

 

[rock.freak667]

Since they are suspended by a light thread, in your free body diagram, included in it should be two tension forces. So you should be able to split that into x and y components and solve.

 

[Kudo Shinichi]

Since they are suspended by a light thread, in your free body diagram, included in it should be two tension forces. So you should be able to split that into x and y components and solve.

Sorry, I don't quite understand how to solve it with tension forces

 

[rock.freak667]

Sorry, I don't quite understand how to solve it with tension forces

The force,F=0.15N you found was the force or repulsion.
Do you have a free body diagram to work with?

 

[Kudo Shinichi]

The force,F=0.15N you found was the force or repulsion.
Do you have a free body diagram to work with?

then isn't the tension force equal and opposite to F=0.15N?
Also I have one more question, where should I use the 0.5m (length of the string) in this problem

 

[rock.freak667]

then isn't the tension force equal and opposite to F=0.15N?

No. The force you calculated is the electric force of repulsion.

Also I have one more question, where should I use the 0.5m (length of the string) in this problem

When you draw the free body diagram, 0.5m is the length of the string which you will need to find the components of the tension forces.

EDIT: post your free body diagram if you are still confused.

 

[Kudo Shinichi]

No. The force you calculated is the electric force of repulsion.



When you draw the free body diagram, 0.5m is the length of the string which you will need to find the components of the tension forces.

EDIT: post your free body diagram if you are still confused.

Here is my free body diagram:
http://tinypic.com/view.php?pic=2djx5b5&s=5"

 

[rock.freak667]

You drew in all of the dimensions correctly, with the weights of masses where it was supposed to be. Now since the force F(=0.15N) is replusive, it should act to move the charges apart. I included another FBD without the lengths associated but only with all the forces acting.

http://img140.imageshack.us/img140/5930/fbdkr2.jpg


Now, since the lengths of both strings are 0.5m, if draw an imaginary line joining the two charges, so that this line, and the lines for the strings, form a triangle, what kind of triangle is made and can you use it to find the angle made by the imaginary line and either of the strings?

 

[Kudo Shinichi]

You drew in all of the dimensions correctly, with the weights of masses where it was supposed to be. Now since the force F(=0.15N) is replusive, it should act to move the charges apart. I included another FBD without the lengths associated but only with all the forces acting.

Now, since the lengths of both strings are 0.5m, if draw an imaginary line joining the two charges, so that this line, and the lines for the strings, form a triangle, what kind of triangle is made and can you use it to find the angle made by the imaginary line and either of the strings?

It's an isosceles triangle
the angles are(64,58,58)
Do I get the force by F=Fsinx?

 

[rock.freak667]

It's an isosceles triangle
the angles are(64,58,58)
Do I get the force by F=Fsinx?

How did you get those angles?

But yes, when you resolve it, you'd see the the x component of T is equal to the force of repulsion. So you can find T.
Then the y component of T is equal to the weight of the charge.

 

[Kudo Shinichi]

How did you get those angles?

But yes, when you resolve it, you'd see the the x component of T is equal to the force of repulsion. So you can find T.
Then the y component of T is equal to the weight of the charge.

I got the angles by drawing a line between the isosceles triangle and use sinx=3/5

sorry, I have one more question, for F=fsinx, f equals to 0.15N right?

 

[rock.freak667]

I got the angles by drawing a line between the isosceles triangle and use sinx=3/5

sorry, I have one more question, for F=fsinx, f equals to 0.15N right?

Check that again, cosx would be 3/5. (adjacent/hypotenuse)
F=fcosx