(a) Consider two equal point charges q, separated by a distance 2a. Construct the plane equidistant from the two charges. By integrating Maxwell’s stress tensor over this plane, determine the force of one charge on the other.
(b) Do the same for charges that are opposite in sign.
(a) (\overleftrightarrow{T} \cdot d a )_{z}=T_{z x} d a_{x}+T_{z y} d a_{y}+T_{z z} d a_{z}.
But for the x y plane d a_{x}=d a_{y}=0, \text { and } d a_{z}=-r d r d \phi (I’ll calculate the force on the upper charge).
(\overleftrightarrow{T} \cdot d a )_{z}=\epsilon_{0}\left(E_{z} E_{z}-\frac{1}{2} E^{2}\right)(-r d r d \phi).
\text { Now } E =\frac{1}{4 \pi \epsilon_{0}} 2 \frac{q}{ᴫ^{2}} \cos \theta \hat{ r }, \text { and } \cos \theta=\frac{r}{ᴫ}, \text { so } E_{z}=0, E^{2}=\left(\frac{q}{2 \pi \epsilon_{0}}\right)^{2} \frac{r^{2}}{\left(r^{2}+a^{2}\right)^{3}} . Therefore
F_{z}=\frac{1}{2} \epsilon_{0}\left(\frac{q}{2 \pi \epsilon_{0}}\right)^{2} 2 \pi \int_{0}^{\infty} \frac{r^{3} d r}{\left(r^{2}+a^{2}\right)^{3}}=\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{2} \int_{0}^{\infty} \frac{u d u}{\left(u+a^{2}\right)^{3}}\left(\text { letting } u \equiv r^{2}\right)
=\left.\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{2}\left[-\frac{1}{\left(u+a^{2}\right)}+\frac{a^{2}}{2\left(u+a^{2}\right)^{3}}\right]\right|_{0} ^{\infty}=\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{2}\left[0+\frac{1}{a^{2}}-\frac{a^{2}}{2 a^{4}}\right]=\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{(2 a)^{2}}.
(b) In this case E =-\frac{1}{4 \pi \epsilon_{0}} 2 \frac{q}{ᴫ^{2}} \sin \theta \hat{ z }, \text { and } \sin \theta=\frac{a}{ᴫ} , so
E^{2}=E_{z}^{2}=\left(\frac{q a}{2 \pi \epsilon_{0}}\right)^{2} \frac{1}{\left(r^{2}+a^{2}\right)^{3}}, \text { and hence }(\overleftrightarrow{T} \cdot d a )_{z}=-\frac{\epsilon_{0}}{2}\left(\frac{q a}{2 \pi \epsilon_{0}}\right)^{2} \frac{r d r d \phi}{\left(r^{2}+a^{2}\right)^{3}} . Therefore
F_{z}=-\frac{\epsilon_{0}}{2}\left(\frac{q a}{2 \pi \epsilon_{0}}\right)^{2} 2 \pi \int_{0}^{\infty} \frac{r d r}{\left(r^{2}+a^{2}\right)^{3}}=-\frac{q^{2} a^{2}}{4 \pi \epsilon_{0}}\left[-\frac{1}{4} \frac{1}{\left(r^{2}+a^{2}\right)^{2}}\right]_{0}^{\infty}=-\frac{q^{2} a^{2}}{4 \pi \epsilon_{0}}\left[0+\frac{1}{4 a^{4}}\right]=-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{(2 a)^{2}} .
