Electric potential energy in $1+1$-dimensional space-time

Question

In $1+1$-dimensional space-time, Gauss's law implies that

$$\int\ \vec{E}\cdot{d\vec{A}}=\displaystyle{\frac{Q}{\epsilon_{0}}} \implies 2 E =\displaystyle{\frac{Q}{\epsilon_{0}}} \implies E =\displaystyle{\frac{Q}{2\epsilon_{0}}},$$

where the factor of $2$ comes from the two endpoints of the Gaussian 'surface' with the charge $Q$ at the centre.

So, $$V=-\int\ \vec{E}\cdot{d\vec{r}} \sim -Qx,$$

where $x$ is the distance from the charge $Q$ and hence is necessarily non-negative.

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Now, consider the charge configuration where two massive charges $+Q$ are separated by a distance $d$ and a light charge $-q$ oscillates in between the two massive charges. The light charge $-q$ is attached to one of the massive charges $+Q$ via a spring which causes the oscillation of the light charge $-q$.

So, $$V(x) \sim x^{2} + |x|,$$

where $x$ is the displacement from the equilibrium position.

Why is the electric potential energy a function of the absolute value of the displacement of the light charge $-q$ ?

Answer 1

In this answer we would like to clarify the formulation of E&M in 1+1D with point charges $q_1,\ldots, q_n$, at positions $x_1(t),\ldots, x_n(t)$. The charge density is $$ \rho(x,t)~=~\sum_{i=1}^n q_i \delta(x\!-\!x_i(t)). \tag{1}$$ The total charge in the interval $[a,b]$ is $$ Q([a,b]) ~=~\int_{[a,b]} \!dx ~\rho~=~\epsilon_0\Phi_E, ~\qquad \Phi_E~=~E(x\!=\!b)-E(x\!=\!a), \tag{2} $$ which is Gauss's law in integral form. A note about sign conventions: Gauss's law measures the electric flux $\Phi_E$ out of the interval $[a,b]$. The electric field $E$ is measured positive in the positive $x$-direction. Therefore the contribution to the electric flux is $-E(x\!=\!a)$ at the lower endpoint $x\!=\!a$.

Gauss's law in differential form reads $$ \frac{\partial E}{\partial x}~=~\frac{\rho}{\epsilon_0}. \tag{3}$$ The electric field is $$ E(x,t)~=~\sum_{i=1}^n \frac{q_i}{2\epsilon_0} {\rm sgn}(x\!-\!x_i(t))~=~-\frac{\partial V(x,t)}{\partial x}. \tag{4}$$ The potential field is $$ V(x,t)~=~-\sum_{i=1}^n \frac{q_i}{2\epsilon_0} |x\!-\!x_i(t)|. \tag{5}$$ There is no magnetic field. See also e.g. my Phys.SE answers here and here.