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Coulomb's Law

The force between two charges, q1,q2q_1, q_2 at a distance r12r_{12} is given by the Coulomb's law:

Fe=ke∣q1∣∣q2∣r2,\begin{equation}F_e = k_e \frac{|q_1| |q_2|}{r^2},\end{equation}

where:

ke=14πϵ0≈9⋅109 Nm2C−2.\begin{equation}k_e = \frac{1}{4 \pi \epsilon_0} \approx 9 \cdot 10^9\ N m^2 C^{-2}.\end{equation}

If the product q1q2q_1 q_2 is positive, the force is repulsive. If the product is negative, the force is attractive.

Vector form:

Fe=keq1q2r122r^12,\begin{equation}\boldsymbol{F}_e = k_e \frac{q_1 q_2}{r_{12}^2} \boldsymbol{\hat{r}_{12}},\end{equation}

where:

r12=r1−r2,r^12=r12r12. \begin{align*} \boldsymbol{r}_{12} &= \boldsymbol{r}_1 - \boldsymbol{r}_2, \\ \boldsymbol{\hat{r}_{12}} &= \frac{\boldsymbol{r}_{12}}{r_{12}}. \end{align*}

The unit of charge is coulomb (CC)

In a non-vacuum environment, we have to use the relative permittivity, ϵr\epsilon_r:

Fe=keϵr∣q1∣∣q2∣r2,Fe=keϵrq1q2r122r^12. \begin{align} F_e &= \frac{k_e}{\epsilon_r} \frac{|q_1| |q_2|}{r^2}, \\ \boldsymbol{F}_e &= \frac{k_e}{\epsilon_r} \frac{q_1 q_2}{r_{12}^2} \boldsymbol{\hat{r}_{12}}. \end{align}
MaterialRelative permittivity
Vacuum1
Air≈1\approx 1
Water81,6

The smallest charge, the elementary charge ee is equal to:

e≈1.602⋅10−19 C.e \approx 1.602 \cdot 10^{-19}\ C.

Two balls carrying charges q1=−20 nCq_1 = -20\ nC and q2=80 nCq_2 = 80\ nC are placed at a distance of r12=10 cmr_{12} = 10\ cm apart. What is the magnitude of the force they act on each other? What is the magnitude of the force if they are placed in water?

When placed in vacuum:

q1=−2⋅10−8 C,q2=8⋅10−8 C,r=0.1 m,Fe=ke∣q1∣∣q2∣r2=9⋅1092−8⋅8−80.12=1.44⋅10−3 N. \begin{align*} q_1 &= -2 \cdot 10^{-8}\ C, \\ q_2 &= 8 \cdot 10^{-8}\ C, \\ r &= 0.1\ m, \\ F_e &= k_e \frac{|q_1| |q_2|}{r^2} \\ &= 9 \cdot 10^9 \frac{2^{-8} \cdot 8^{-8}}{0.1^2} \\ &= 1.44 \cdot 10^{-3}\ N. \end{align*}

When placed in water:

q1=−2⋅10−8 C,q2=8⋅10−8 C,r=0.1 m,ϵr=81.6,Fe=keϵr∣q1∣∣q2∣r2=9⋅10981.62−8⋅8−80.12≈1.76⋅10−5 N. \begin{align*} q_1 &= -2 \cdot 10^{-8}\ C, \\ q_2 &= 8 \cdot 10^{-8}\ C, \\ r &= 0.1\ m, \\ \epsilon_r &= 81.6, \\ F_e &= \frac{k_e}{\epsilon_r} \frac{|q_1| |q_2|}{r^2} \\ &= \frac{9 \cdot 10^9}{81.6} \frac{2^{-8} \cdot 8^{-8}}{0.1^2} \\ &\approx 1.76 \cdot 10^{-5}\ N. \end{align*}
ϵr=1Fe=1.44⋅10−3 NAttractive. \begin{align*} \epsilon_r &= 1 \\ F_e &= 1.44 \cdot 10^{-3}\ N &\textrm{Attractive}. \end{align*}

A 10 nC10\ nC is placed at the origin and a −30 nC-30\ nC charge is placed 20 cm20\ cm to the right. A third 15 nC15\ nC charge is placed 15 cm15\ cm above the second charge. What is the total force on the second charge due to the other two?

q1=10−8 C,q2=−3⋅10−8 C,q3=2⋅10−8 C,r1=0,r2=0.2 m x^,r3=0.2 m x^+0.15 m y^,r21=r2−r1=0.2 m x^,r21=0.2 m,r^21=x^,r23=r2−r3=−0.15 m y^,r23=0.15 m,r^23=−y^. \begin{align*} q_1 &= 10^{-8}\ C, \\ q_2 &= -3 \cdot 10^{-8}\ C, \\ q_3 &= 2 \cdot 10^{-8} \ C, \\ \boldsymbol{r}_1 &= \boldsymbol{0}, \\ \boldsymbol{r}_2 &= 0.2\ m\ \boldsymbol{\hat{x}}, \\ \boldsymbol{r}_3 &= 0.2\ m\ \boldsymbol{\hat{x}} + 0.15\ m\ \boldsymbol{\hat{y}}, \\[1.5ex] \boldsymbol{r}_{21} &= \boldsymbol{r}_2 - \boldsymbol{r}_1 \\ &= 0.2\ m\ \boldsymbol{\hat{x}}, \\ r_{21} &= 0.2\ m, \\ \boldsymbol{\hat{r}_{21}} &= \boldsymbol{\hat{x}}, \\[1.5ex] \boldsymbol{r}_{23} &= \boldsymbol{r}_2 - \boldsymbol{r}_3 \\ &= -0.15\ m\ \boldsymbol{\hat{y}}, \\ r_{23} &= 0.15\ m, \\ \boldsymbol{\hat{r}_{23}} &= -\boldsymbol{\hat{y}}. \end{align*}

The force on second charge due to the first one:

Fe1=keq2q1r212r^21=9⋅109−3⋅10−8⋅10−80.22x^=−6.75⋅10−5 m x^. \begin{align*} \boldsymbol{F}_{e1} &= k_e \frac{q_2 q_1}{r_{21}^2} \boldsymbol{\hat{r}_{21}} \\ &= 9 \cdot 10^9 \frac{-3 \cdot 10^{-8} \cdot 10^{-8}}{0.2^2} \boldsymbol{\hat{x}} \\ &= -6.75 \cdot 10^{-5}\ m\ \boldsymbol{\hat{x}}. \end{align*}

The force on second charge due to the third one:

Fe3=keq2q3r232r^23=9⋅109−3⋅10−8⋅2⋅10−80.152(−y^)=2.4⋅10−4 m y^. \begin{align*} \boldsymbol{F}_{e3} &= k_e \frac{q_2 q_3}{r_{23}^2} \boldsymbol{\hat{r}_{23}} \\ &= 9 \cdot 10^9 \frac{- 3 \cdot 10^{-8} \cdot 2 \cdot 10^{-8}}{0.15^2} (- \boldsymbol{\hat{y}}) \\ &= 2.4 \cdot 10^{-4}\ m\ \boldsymbol{\hat{y}}. \end{align*}

The total force is just sum of these two forces:

Fe=Fe1+Fe3=−6.75⋅10−5 m x^+2.4⋅10−4 m y^. \begin{align*} \boldsymbol{F}_e &= \boldsymbol{F}_{e1} + \boldsymbol{F}_{e3} \\ &= -6.75 \cdot 10^{-5}\ m\ \boldsymbol{\hat{x}} + 2.4 \cdot 10^{-4}\ m\ \boldsymbol{\hat{y}}. \end{align*}

Charge density is an amount of electric charge per unit of length, surface area or volume.

For a small change in length, dldl and a small change in charge dqdq, the linear charge density, λ\lambda, is equal to:

λ=dqdl.\lambda = \frac{dq}{dl}.

For a small change in surface area, dSdS and a small change in charge dqdq, the surface charge density, σ\sigma, is equal to:

σ=dqdS.\sigma = \frac{dq}{dS}.

For a small change in volume, dVdV and a small change in charge dqdq, the volume charge density, ρ\rho, is equal to:

ρ=dqdV.\rho = \frac{dq}{dV}.

For constant charge densities, the equations simplify to:

λ=ql,σ=qS,ρ=qV. \begin{align*} \lambda &= \frac{q}{l}, \\ \sigma &= \frac{q}{S}, \\ \rho &= \frac{q}{V}. \end{align*}