Question #98f20

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Answer 1 · 2015-11-25T14:10:17

#q=6.66xx10^(-9)"C"#

MFDocs

The weight acting downwards is balanced by the vertical component of the tension.

The electrostatic repulsion between the 2 charges is balanced by the horizontal component of the tension.

From the geometry of the triangle we can write:

#sin15=r/0.45#

#:.r=0.45xx0.259=0.11633#

#:.2r=0.233"m"#

Taking the vertical components:

#mg=Tcos15#

#:.T=(mg)/(cos15)=(2.8xx10^(-6)xx9.8)/(0.966)#

#:.T=28.405"N"#

Now for the horizontal components:

The force between the 2 charges is given by Coulomb's Law:

#F=(1)/(4piepsilon_0).(q^2)/r^2#

#(1)/(4piepsilon_0)# is a constant and has the value #k=9xx10^(9)"m/F"#

#:.F_E=(kq^2)/(0.233^2)=Tcos75#

Substituting in the value for #T#:

#:.(kq^2)/0.233^2=28.405xx10^(-6)xx0.259#

#:.q^2=(28.405xx10^(-6)xx0.259xx0.233^2)/(9xx10^9)#

#q^2=(0.4xx10^(-6))/(9xx10^9)#

#q^2=0.444xx10^(-16)#

#q=6.66xx10^(-9)"C"#