- #1
NotaPhysicsMan
- 146
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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!
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!