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March 29, 2015

March 29, 2015

Posted by **Hannah (Please Help)** on Tuesday, June 5, 2012 at 3:32pm.

- physics -
**Elena**, Tuesday, June 5, 2012 at 4:10pmIf the charged ball is suspended be the string which is deflected by the angle α, the forces acting on it are: mg (downwards), tension T (along the string - to the pivot point), and F (electric force –along the line connecting the charges).

Projections on the horizontal and vertical axes are:

x: T•sin α = F, ….(1)

y: T•cosα = mg. ….(2)

Divide (1) by (2):

T•sin α/ T•cosα = F/mg,

tan α = F/mg.

Since

q1=q2=q.

r=2•L•sinα,

k=9•10^9 N•m²/C²

F =k•q1•q2/r² = k•q²/(2•L•sinα)².

tan α = F/mg =

= k•q²/(2•L•sinα)² •mg.

q = (2•L•sinα) • sqrt(m•g•tanα/k)=

=(2•1.7•0.5) •sqrt(0.0112•9.8•0.577/9•10^9) =

=4.5•10^-6 C =4.5 μC.

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