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Physics Question on coulombs law

Two small charged spheres have charges of 2×10−7 C2 \times 10^{-7}\, C and 3×10−7C3 \times 10^{-7}C . They are placed 30 cm30\,cm apart in air. What is the force between them ?

A

F=6×10−3 NF=6 \times 10^{-3}\,N

B

F=0.6×10−3 NF=0.6 \times 10^{-3}\,N

C

F=6.5×10−3 NF=6.5 \times 10^{-3}\,N

D

F=0.65×10−3 NF=0.65 \times 10^{-3}\,N

Answer

F=6×10−3 NF=6 \times 10^{-3}\,N

Explanation

Solution

q1=2×10−7C;q_{1}=2\times 10^{-7}C;
q2=3×10−7 Cq_{2}=3\times 10^{-7}\,C
r=30 cmr=30\,cm
=0.3 m;=0.3\,m;
F=q1q24πε0r2F=\frac{q_{1}q_{2}}{4 \pi \varepsilon_{0} r^{2}}
where ε0 \varepsilon_{0} = Permittivity o f free space
14πε0=9×109Nm2C−2\frac{1}{4\pi\varepsilon_{0}}=9\times10^{9} Nm^{2}C^{-2}
F=9×109×2×10−7×3×10−7(0.3)2F=\frac{9\times10^{9}\times2\times10^{-7}\times3\times10^{-7}}{\left(0.3\right)^{2}}
=6×10−3N=6\times10^{-3}N