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Q.The vector form of Coulomb's law is (A) F = (1/(4πε_0)) · (q_1 q_2/|r|^3) r
(B) F = (1/(4πε_0)) · (q_1 q_2/|r|^3)
(C) F = (1/(4πε_0)) · (q_1 q_2/r^2) r
(D) F = (1/(4πε_0)) · (q_1 q_2/|r|^2) r

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F⃗=14πε0q1q2∣r⃗∣3 r⃗\vec{F} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{|\vec r|^{3}}\,\vec r.

Coulomb's law in vector form must give the force magnitude ∝1/r2\propto 1/r^2 and point along the line joining the charges. Writing the displacement vector as r⃗\vec r (magnitude rr) and using the unit vector r^=r⃗/∣r⃗∣\hat r = \vec r/|\vec r|:

F⃗=14πε0q1q2r2 r^=14πε0q1q2∣r⃗∣3 r⃗.\vec{F} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\,\hat r = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{|\vec r|^{3}}\,\vec r.

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