So Coulomb's law calculates the electric force *F* in newtons
(N) between two electric charges
*q*_{1} and *q*_{2} in coulombs (C)

with a distance of *r* in meters (m).

*F* is the force on *q*_{1}
and *q*_{2} measured in newtons (N).

*k* is Coulomb's constant *k* =
8.988×10^{9} N⋅m^{2}/C^{2}

*q*_{1} is the first charge in
coulombs (C).

*q*_{2} is the second charge in
coulombs (C).

*r* is the distance between the 2 charges
in meters (m).

So When charges q1 and q2 is increased, the force F is increased.

So When distance r is increased, the force F is decreased.

So Find the force between 2 electric charges of 2×10^{-5}C
and 3×10^{-5}C with distance of
40cm between them.

*q*_{1 }= 2×10^{-5}C

*q*_{2 }= 3×10^{-5}C

*r *= 40cm = 0.4m

*F* = k×q_{1}×q_{2
}/ r^{2} = 8.988×10^{9}N⋅m^{2}/C^{2}
× 2×10^{-5}C
× 3×10^{-5}C / (0.4m)^{2}
= 37.705N

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