Asked by Sandhya
A thin insulating thread of length L is attached to a point on the perimeter of a thin disk.
Disk is has radius R< L and is charged with a uniform surface density, σ>0. At the end of
the thread there is a small ball mass m and charge q>0. The mass happens to be located
on disk's axis. What is the mass of the ball?
Disk is has radius R< L and is charged with a uniform surface density, σ>0. At the end of
the thread there is a small ball mass m and charge q>0. The mass happens to be located
on disk's axis. What is the mass of the ball?
Answers
Answered by
bobpursley
Figure the Electric field from the charged disk along the axis at a distance z, given R
http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elelin.html#c3
Now, given that expression, then you know from the geometry with the angle theta the string makes from disk,
tanTheta=Eq/mg
But CosineTheta=R/L
and sinTheta=sqrt((L^2-R^2)/L^2) or
sinTheta=sqrt(1-(R/L)^2)
but sinTheta/cosTheta= tan Theta, so
Eq/mg=sqrt(1-(R/L)^2) *L/R
Eq/mg=sqrt((L/R)^2-1)
and now you can solve for L
Remember in your E expression, z=Sqrt(L^2-R^2)
As you can tell, some algebra will be required. Get a nice pad of paper.
http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elelin.html#c3
Now, given that expression, then you know from the geometry with the angle theta the string makes from disk,
tanTheta=Eq/mg
But CosineTheta=R/L
and sinTheta=sqrt((L^2-R^2)/L^2) or
sinTheta=sqrt(1-(R/L)^2)
but sinTheta/cosTheta= tan Theta, so
Eq/mg=sqrt(1-(R/L)^2) *L/R
Eq/mg=sqrt((L/R)^2-1)
and now you can solve for L
Remember in your E expression, z=Sqrt(L^2-R^2)
As you can tell, some algebra will be required. Get a nice pad of paper.
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