How Do You Calculate Charge and Tension in Hanging Charged Balls?

[NotaPhysicsMan]

So here's the question:

Two identical small insulating balls are suspended by separate 0.25m threads that are attached to a common point on the ceiling. Each ball has a mass of 8.0 x10^-4kg. Initially the balls are unchaged and hang straight down. They are then given identical positive charges and as a result, spread apart with an angle of 36 degrees between the threads. Determine a) the charge on each ball and b) the tension in the threads.

Ok, here's my work, check please!

a)ok I'll separte into components.

Fnetx=-F(electric force)+Fe
0=-K|q1||q2|/r1^2 + ( K|q1||q2|/r2^2). Both these cancel so =0

Fnety=F(electric)y+Fey
mg=-K|q1||q2|/r1^2 (cos18degrees) + [( -K|q1||q2|/r2^2) x (cos18degrees)]
8.0x10^-4kg x 9.81=-2k|q1||q2|/r^2

so I now isolate for the q1 and q2 since they're the same and solve for them.
|q1||q2|=2.87x10^-14 C
each charge then has 1.43x10^-14 C

B)the tension is just T=mg

so, separate into components again!

x1=T1cos18
y2=T1sin18

x2=T2cos18
y2=T2sin18.

X's cancel.

Y is Fnet=T1sin18+T2sin18

or mg=2Tsin18

solve for T=mg/2sin18
T=0.0127 N??

Thanks for you help!

 

[NotaPhysicsMan]

Must be a slow night *Bump*

 

[MathStudent]

So here's the question:

Two identical small insulating balls are suspended by separate 0.25m threads that are attached to a common point on the ceiling. Each ball has a mass of 8.0 x10^-4kg. Initially the balls are unchaged and hang straight down. They are then given identical positive charges and as a result, spread apart with an angle of 36 degrees between the threads. Determine a) the charge on each ball and b) the tension in the threads.

Ok, here's my work, check please!

a)ok I'll separte into components.

Fnetx=-F(electric force)+Fe
0=-K|q1||q2|/r1^2 + ( K|q1||q2|/r2^2). Both these cancel so =0

Fnety=F(electric)y+Fey
mg=-K|q1||q2|/r1^2 (cos18degrees) + [( -K|q1||q2|/r2^2) x (cos18degrees)]
8.0x10^-4kg x 9.81=-2k|q1||q2|/r^2

so I now isolate for the q1 and q2 since they're the same and solve for them.
|q1||q2|=2.87x10^-14 C
each charge then has 1.43x10^-14 C

Ok its pretty much all wrong from the start...

I'm going to assume throughout that the sphere's charge is uniformly distributed, because if it's not then this problem isn't given enough info.

To start draw a FBD for ONE of the spheres and indicate all the forces ACTING on it. There should only be ONE electrical force, the force from the other sphere, (The charges on the sphere your analyzing doesn't effect itself).
There should also be a Tension force and the Weight of the sphere.

Let the horizontal axis of the FBD be horizontal wrt the ceiling and let it intersect the spheres center.

The force of tension from the string pulls on the sphere, it doesn't push, so its vector representation should be directed away from the sphere and towards the point of contact with the ceiling.

What is the angle that this tension force makes with the HORIZONTAL axis, ( it isn't 18 degrees - that's the angle it makes wrt the vertical axis ).

Set up your Newtons 2nd law equations for the vertical and horizontal components of force and note that this system is in static equillibrium.

hint: it will probably be easier to solve part B first but that uses the same guidelines as above.

B)the tension is just T=mg

so, separate into components again!

x1=T1cos18
y2=T1sin18

x2=T2cos18
y2=T2sin18.

X's cancel.

Y is Fnet=T1sin18+T2sin18

or mg=2Tsin18

solve for T=mg/2sin18
T=0.0127 N??

Thanks for you help!

I'm pretty sure they want the force of tension when the balls are charged and repelling each other, otherwise it would be too easy.

Follow the guidlines I gave above. It will probably be easier to solve part B before part A.

come back if you run into any more trouble

-MS

 

[MathStudent]

see if you can set up the Newtons 2nd law equations correctly before carrying on, as the rest of this problem is dependent upopn these.

 

[NotaPhysicsMan]

So let me see.

I see three principle forces. One is tension, one is the electric force and the other gravity.

Fnetx=Felectric-Tsin72degrees=0
Fnety=mg-Tcos72degrees=0

Where it doesnt' make sense is that I have two unknowns for Fnetx. One is the q1q2 and the other tension.

 

[MathStudent]

So let me see.

I see three principle forces. One is tension, one is the electric force and the other gravity.

Fnetx=Felectric-Tsin72degrees=0
Fnety=mg-Tcos72degrees=0

Where it doesnt' make sense is that I have two unknowns for Fnetx. One is the q1q2 and the other tension.

you have the sin and cos mixed up,, remember cos is adj/hyp

\Sigma \vec{F_x} = \vec{F_E} - T cos(72) = 0

You can use your \Sigma \vec{F_y} equation to solve for T since both the m and g are known right?

Then you can use that value of T to solve for \vec{F_E} in the \Sigma \vec{F_x} equation.

 

[NotaPhysicsMan]

wait, I switched it around...

Fnetx=Felectric-Tcos72 degrees=0
Fnety=mg-Tsin72degrees=0.

From this I'm gussing I can sovle T=mg/sin72?
T=8.25x10^-3 N

Ok now I can solve for Felectric.

0=Fe-Tcos72
=k(q1)(q2)/r^2
Tcos72=k(q1)(q2)/r^2

q1q2=Tcos72 x r^2/(k)

=8.25x10^-3 cos 72 x 0.25^2/(8.99x10^9)
ok so q1q2=1.773 x 10^-14 C?

divide by 2 , then each charge is 8.86x10^-15C?

 

[NotaPhysicsMan]

lol I can't see your latex generated formulas! maybe something wrong with Java...

 

[MathStudent]

wait, I switched it around...

Fnetx=Felectric-Tcos72 degrees=0
Fnety=mg-Tsin72degrees=0.

From this I'm gussing I can sovle T=mg/sin72?
T=8.25x10^-3 N

right!

Ok now I can solve for Felectric.

0=Fe-Tcos72
=k(q1)(q2)/r^2
Tcos72=k(q1)(q2)/r^2

q1q2=Tcos72 x r^2/(k)

=8.25x10^-3 cos 72 x 0.25^2/(8.99x10^9)

r is the distance between the two spheres, which doesn't equal the length of the string.

ok so q1q2=1.773 x 10^-14 C?

divide by 2 , then each charge is 8.86x10^-15C?

the problem said that each sphere is given equal charge so in this case
q1 = q2
therefore q1q2 = q^2
so you need to take the square root, not divide by 2

 

[NotaPhysicsMan]

Oh right yes, so the r is then just 0.25sin18degrees x 2=0.1545m

so q1q2 then according to you, take the root of 6.769x10^-13 = 2.601x10^-13?

Let this be right lol.

 

[MathStudent]

Oh right yes, so the r is then just 0.25sin18degrees x 2=0.1545m

so q1q2 then according to you, take the root of 6.769x10^-13 = 2.601x10^-13?

Let this be right lol.

not quite, the 0.25sin(18) just calculates half the distance between them ( do you see this ) so you need to multiply this by 2. And the square root of 6.769*10^-13 is not 2.60*10^-13,, did you use a calculator?

 

[NotaPhysicsMan]

I did I did lol. 0.25sin18=0.077 x 2 =0.155.

 

[MathStudent]

I did I did lol. 0.25sin18=0.077 x 2 =0.155.

oh sorry, I overlooked that... the answer is still incorrect. However all your values are right, so it must be a calculational error,,,, show what you put in the calculator

 

[NotaPhysicsMan]

hmm ok forgot brackets, the answer is then 8.22x10^-7C!

 

[MathStudent]

I got 8.22*10^-8 C, your off by a factor of 10 somewhere

I input q = 0.50*sin(18)*sqrt( 8.25*10^-3 * cos(72) / 9*10^9 )

 

[NotaPhysicsMan]

No way, my cheesy scientific calc says -7 . I dont' know, I'll try it on a diff one. But thanks either way!

 

[MathStudent]

well you've gotten the right calculations so far, so its probably more likely to be user error than calculator error ...
but good luck to you, glad I could help
-MS