Coulomb's law for continuous charge distributions - Exercises

1. The potential \(\d V\) at the point \((0,0,r)\) due to a small length \(\d x\) of the line is given by $$\begin{equation} \d V = \frac{\d Q}{4\pi\epsilon_0 R} = \frac{\lambda \d x}{4\pi\epsilon_0\sqrt{x^2+r^2}}\end{equation}$$ Thus $$\begin{equation} V = \int_{-\ell_1}^{\ell_2} \frac{\lambda \d x}{4\pi\epsilon_0\sqrt{x^2+r^2}} = \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2 +\sqrt{\ell_2^2+r^2}}{-\ell_1 + \sqrt{\ell_1^2+r^2}}\right| \end{equation}$$
2. If \(r \gg \ell_1+\ell_2\), then we expect the line of charge to behave as a point charge with magnitude \(Q = \lambda(\ell_1+\ell_2)\). That would be \(\displaystyle{V \approx \frac{\lambda(\ell_1+\ell_2)}{4\pi\epsilon_0 r}}\). Recall \(\ln(1+x) \approx x\) for small \(x\). For the formula from part (a) we have $$\begin{align} V ={}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2 +\sqrt{\ell_2^2+r^2}}{-\ell_1 + \sqrt{\ell_1^2+r^2}}\right|\\ \approx{}& \frac{\lambda}{4\pi\epsilon_0}\ln\left|\frac{\ell_2 + r}{r - \ell_1}\right| \\ ={}& \frac{\lambda}{4\pi\epsilon_0}\ln\left|\frac{\ell_2 +\ell_1 + r - \ell_1}{r - \ell_1}\right|\\ ={}& \frac{\lambda}{4\pi\epsilon_0}\ln\left|1 + \frac{\ell_2 +\ell_1}{r - \ell_1}\right|\\ \approx{}& \frac{\lambda}{4\pi\epsilon_0} \frac{\ell_1 +\ell_2}{r - \ell_1}\\ \approx{}& \frac{\lambda}{4\pi\epsilon_0} \frac{\ell_1 +\ell_2}{r} \end{align}$$ Which agrees with what we have from the potential formula as if the length of charge was a point charge.
3. Recall that \((x+1)^{1/2} \approx 1+x/2\) for small \(x\). If \(r \ll \ell_1 + \ell_2\), then $$\begin{align} V ={}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2 +\sqrt{\ell_2^2+r^2}}{-\ell_1 + \sqrt{\ell_1^2+r^2}}\right|\\ ={}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2(1 +\sqrt{1+r^2/\ell_2^2})}{\ell_1(-1 + \sqrt{1+r^2/\ell_1^2})}\right|\\ \approx{}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2(1 +1+r^2/2\ell_2^2)}{\ell_1(-1 +1+r^2/2\ell_1^2)}\right|\\ ={}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2(2+r^2/2\ell_2^2)}{\ell_1(r^2/2\ell_1^2)}\right|\\ \approx{}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{\ell_2(2)}{r^2/2\ell_1}\right|\\ ={}& \frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{4\ell_1\ell_2}{r^2}\right|\\ \end{align}$$ Differentiating, $$\begin{align} \mathbf{E} ={}& \left(-\frac{\d}{\d r}V\right)\rhat\\ ={}&\left(-\frac{\d}{\d r} \left(\frac{\lambda}{4\pi\epsilon_0}\ln \left| \frac{4\ell_1\ell_2}{r^2}\right|\right)\right)\rhat\\ ={}& \frac{\lambda}{2\pi\epsilon_0 r}\rhat \end{align}$$ Using Gauss's law with a cylindrical gaussian surface, with axis along the line gives us $$\begin{equation} \mathbf{E} = \frac{\lambda}{2\pi\epsilon_0 r}\rhat \end{equation}$$ Which agrees with the above result.

To find the force on charge \(q\) we will integrate the force over the circular arc. Linear charge density is $$\begin{equation}\lambda = \frac{Q}{a\alpha}\end{equation}$$ Consider the force \(\d\mathbf{F}\) due to a small length of charge \(\d Q\), which resides at an angle \(\theta\). In order to simplify the problem, pretend like the arc extends along \([-\alpha/2,\alpha/2]\) instead of \([0,\alpha]\). This will cause the \(y\) components of the force to cancel out, so we would only need to find the force in the \(x\) direction. The force will then be $$\begin{equation} \d F_x = \frac{1}{4\pi\epsilon_0} \frac{q\d Q\cos\theta}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{q \cos\theta \d Q }{a^2} \end{equation}$$ Since \(\d Q = \lambda a\d\theta\), we can plug that in to the resulting integral: $$\begin{equation} F_x = \frac{1}{4 \pi \epsilon _0} \int_{-\alpha/2}^{\alpha/2}\frac{q\lambda}{a}\cos\theta \d\theta=\frac{1}{4 \pi \epsilon _0} 2\int_{0}^{\alpha/2}\frac{ qQ}{\alpha a^2}\cos\theta \d\theta \end{equation}$$ Thus we have for the magnitude of the force in the \(x\) direction (with the rotated arc) to be $$\begin{equation} F_x =\frac{1}{4 \pi \epsilon _0} \frac{qQ}{a^2 \alpha}2\sin\left(\frac{\alpha}{2}\right) = \frac{1}{2 \pi \epsilon _0} \frac{qQ}{a^2 \alpha}\sin\left(\frac{\alpha}{2}\right) \end{equation}$$ If we want to find the force on the charge \(q\) when the arc is in its true position, we can deduce from symmetry that resultant force will be at an angle of \(\alpha/2\), and pointing towards the origin. Thus we can multiply the magnitude of the force by a unit vector that points towards the origin. That unit vector will be $$\begin{equation} \mathbf{u} = \left(-\cos \left(\frac{\alpha}{2}\right) ,-\sin \left(\frac{\alpha}{2}\right)\right) \end{equation}$$ Thus the force on the charge \(q\) due to the wire which arcs from \([0,\alpha]\) will be $$\begin{equation} \mathbf{F} = \frac{1}{2 \pi \epsilon _0} \frac{qQ}{a^2 \alpha}\sin\left(\frac{\alpha}{2}\right) \left(-\cos \left(\frac{\alpha}{2}\right) ,-\sin \left(\frac{\alpha}{2}\right)\right) \end{equation}$$

1. Let \(\lambda\) be the linear charge density of the square insulator. Then, a little element of charge on the insulator is given by \(\d Q = \lambda \d x\), and the distance from any point \((x,y)\) on the square to the charge is given by \(r = \sqrt{x^2 + y^2 + \ell^2}\). First, note that the \(x\) and \(y\) components of the force on \(Q\) will cancel out, and it will only be pushed upwards. Thus, we only need to find the \(z\) component of force on \(Q\) from one of the sides, then multiply by 4. Let's integrate along the \(x\) direction. $$\begin{align} \d F_z ={}& \frac{Q\d Q\cos\theta}{4\pi\epsilon_0 r^2} \\ ={}& \frac{Q\lambda \d x\cos\theta}{4\pi\epsilon_0 r^2} \\ ={}& \frac{Q\lambda \d x}{4\pi\epsilon_0 r^2}\cdot \frac{\ell}{r} \\ ={}& \frac{Q\ell\lambda \d x}{4\pi\epsilon_0 (x^2+y^2+\ell^2)^{3/2}} \end{align}$$ Thus the force on \(Q\) is given by \(\mathbf{F} = F_z \zhat\). Note that we set \(y = a/2\) because we integrate along one side of the square, then multiply the final result by 4. $$\begin{align} F_z ={}& \int^{a/2}_{-a/2} \frac{Q\ell\lambda \d x}{4\pi\epsilon_0 (x^2 + y^2 + \ell^2)^{3/2}} \\ ={}& \frac{Q\ell\lambda}{4\pi\epsilon_0}\cdot \int^{a/2}_{-a/2}\frac{\d x}{(x^2+(a/2)^2 + \ell^2)^{3/2}} \\ ={}& \frac{Q\ell\lambda}{\pi\epsilon_0}\cdot \int^{a/2}_{-a/2} \frac{2\d x}{(4x^2+(4\ell^2+a^2))^{3/2}} \end{align}$$ Now let's evaluate the integral. First, we find the antiderivative. Let \(2x = \sqrt{4\ell^2+a^2}\tan\theta \Rightarrow 2\d x = \sqrt{4\ell^2+a^2}\sec^2\theta \d\theta\). Then $$\begin{align} \int \frac{2\d x}{(4x^2+(4\ell^2+a^2))^{3/2}} ={}& \int \frac{\sqrt{4\ell^2+a^2}\sec^2\theta \d\theta}{[(4\ell^2+a^2)\tan^2\theta+(4\ell^2+a^2)]^{3/2}} \\ ={}& \frac{1}{4\ell^2+a^2}\cdot \int \cos\theta \d\theta = \frac{1}{4\ell^2+a^2}\sin\theta \end{align}$$ Expressing \(\sin\theta\) in terms of \(x\) gives $$\begin{equation} \frac{1}{4\ell^2+a^2}\sin\theta = \frac{1}{4\ell^2+a^2}\cdot \frac{2x}{\sqrt{4x^2+4\ell^2+a^2}} \end{equation}$$ Now, plugging this back into the original expression gives: $$\begin{align} \frac{Q\ell\lambda}{\pi\epsilon_0}\cdot \int^{a/2}_{-a/2} \frac{2\d x}{(4x^2+(4\ell^2+a^2))^{3/2}} ={}& \frac{Q\ell\lambda}{\pi\epsilon_0(4\ell^2+a^2)}\cdot \left.\left( \frac{2x}{\sqrt{4x^2+4\ell^2+a^2}}\right)\right|^{a/2}_{-a/2}\\ ={}& \frac{Q\ell a\sqrt{2}\lambda}{\pi\epsilon_0(4\ell^2+a^2)\sqrt{a^2+2\ell^2}} \end{align}$$ Thus the force on the charge (multiplying by 4) \(Q\) is $$\begin{equation} \mathbf{F} = \frac{Q\ell a\lambda4\sqrt{2}}{\pi\epsilon_0(4\ell^2+a^2)\sqrt{a^2+2\ell^2}}\zhat \end{equation}$$
2. If \(\ell \gg a\), then \(4\ell^2 + a^2\approx 4\ell^2\), and \(\sqrt{a^2+2l^2}\approx \sqrt{2}\ell\). Then we have $$\begin{equation} \mathbf{F}\approx \frac{Qa\lambda}{\pi\epsilon_0 \ell^2}\zhat \end{equation}$$ which agrees with the model of approximating the charged square as a point charge of magnitude \(4a\lambda\), at distance \(\ell\) from the charge \(Q\).
3. If \(l \ll a\), then we can approximate \(4\ell^2 + a^2 \approx a^2\), and \(\sqrt{a^2+2\ell^2}\approx a\). Then we have $$\begin{equation} \mathbf{F} \approx \frac{\lambda Q \ell 4\sqrt{2}}{\pi\epsilon_0 a^2} \zhat \end{equation}$$
4. Relabelling \(\ell = z\), we can apply Newton's second law to obtain $$\begin{equation} mz"(t) = \frac{Q\lambda 4\sqrt{2}}{\pi\epsilon_0 a^2}z(t) \end{equation}$$ The angular frequency is then (we have another minus sign inside the square root because \(Q\lambda < 0\), so then we are taking the square root of a positive number) $$\begin{equation} \omega = \sqrt{-\frac{Q\lambda 4\sqrt{2}}{m\pi\epsilon_0 a^2}} \end{equation}$$