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Physics · Ch 10 — Electrostatics

Coulomb's Law in Vector Form

10.4.4

Coulomb's Law in Vector Form

As shown in Fig. 10.4, let q1q_1 and q2q_2 be two point charges situated at points A and B respectively, with r12r_{12} the distance separating them. Coulomb's law can be written precisely in VECTOR form, specifying not just the magnitude but also the exact direction of the force on each charge.

The force F⃗21\vec{F}_{21} exerted ON q2q_2 BY q1q_1 is F⃗21=14πϵ0q1q2r212 r^21,\vec{F}_{21}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{21}^2}\,\hat{r}_{21}, where r^21\hat{r}_{21} is the unit vector directed along AB, away from B. Similarly, the force F⃗12\vec{F}_{12} exerted ON q1q_1 BY q2q_2 is F⃗12=14πϵ0q1q2r122 r^12,\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\,\hat{r}_{12}, where r^12\hat{r}_{12} is the unit vector directed along BA, away from A -- so F⃗12\vec{F}_{12} acts on q1q_1 at A and points along BA, away from A. …

Figure 10.4Fig. 10.4: Coulomb's law in vector form

What this figure shows. Two point charges q1q_1 and q2q_2 located at points A and B respectively, joined by a straight line of length r12r_{12} (the separation between them). Two force vectors are drawn: F⃗21\vec{F}_{21}, the force ON q2q_2 due to q1q_1, drawn at B pointing along the line away from A, with the unit vector r^21\hat{r}_{21} (directed along AB, away from B) used to write F⃗21=14πϵ0q1q2r212r^21\vec{F}_{21}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{21}^2}\hat{r}_{21}; and F⃗12\vec{F}_{12}, the force ON q1q_1 due to q2q_2, drawn at A pointing along the line away from B, with unit vector r^12\hat{r}_{12} (directed along BA, away from A). The figure shows r^12\hat{r}_{12} and r^21\hat{r}_{21} pointing in exactly opposite direct …

Misc Ex.3Example 10.3: Comparing the electrostatic and gravitational forces between two protons

Worked out. For two protons (q1=q2=+1.6×10−19q_1=q_2=+1.6\times10^{-19} C, mass 1.67×10−271.67\times10^{-27} kg each) separated by r=10−15r=10^{-15} m, the electrostatic force is Fe=14πϵ0q2r2=9×109×(1.6×10−19)2(10−15)2≈2.3×102F_e=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}=9\times10^9\times\frac{(1.6\times10^{-19})^2}{(10^{-15})^2}\approx2.3\times10^2 N, while the gravitational force is Fg=Gm2r2=6.674×10−11×(1.67×10−27)2(10−15)2≈1.86×10−34F_g=G\frac{m^2}{r^2}=6.674\times10^{-11}\times\frac{(1.67\times10^{-27})^2}{(10^{-15})^2}\approx1.86\times10^{-34} N, giving a ratio Fe/Fg≈1.23×1036F_e/F_g\approx1.23\times10^{36} -- the electrostatic force is about 36 orders of magnitude stronger than gravity at the same separation. The example draws out both similarities (both obey the inverse-square law, both are central forces acting along the joining line) and differences (gravity is always attractive while the electric force can be attractive or repulsive; gra …