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2015-04-21T18:08:29+05:30

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Say the charge on each ball is q
r=3*10^{-2}m
F=4*10^{-5}N

From coulomb's law we have F=9*10^9\frac{q_1q_2}{r^2}

Plugin the given values and get
4*10^{-5}=9*10^9\frac{q^2}{(3*10^{-2})^2}
q^2=4\times\,10^{-18}
q=\pm 2\times\,10^{-9}C
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what is "r" here
"r" is the distance between charges
in meters
notice, 3cm = 3*10^(-2) m
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2015-04-22T13:40:26+05:30

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The Coulomb's law of force between two electrically charged particles :
         q₁ , q₂ = electric charges on the two particles.
           r = distance between two particles.
      the particles are assumed to be point sized.

Force\ F=\frac{1}{4\pi \epsilon_0}\ \frac{q_1\ q_2}{r^2}\\\\4*10^{-5}N=9*10^9*\frac{q^2}{0.03^2\ m^2}\\\\q^2=\frac{4*10^{-5}\ *\ 0.03^2}{9*10^9}\\\\q=+ 2*10^{-9}C\ \ or\ \ -2*10^{-9}C\\\\q=+2\ nano\ Coulombs\ \ or,\ \ -2\ nC

The charges can either both be positive or negative.

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