Electric field strength with distance

[spaghetti3451]

Consider Gauss's law in $1$ space and $1$ time dimension. In this case,

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

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.

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, $V(x)=\frac{1}{2}kx^{2} + \cdots$ ,

where $x$ is the displacement from the equilibrium position and the dots represent the electric potential energy.

I get a constant electric potential energy (independent of $x$) of the light charge $-q$ due to the two massive charges $+Q$. Do you get the same answer?

 

[andrewkirk]

In 3D we work out the difference in potential between two points under the field of an object at the origin by the integral
$$-\int_a^b \frac{C_3dx}{x^2}$$
where the points are on the same radial line at distances $a$ and $b$.
This gives
$$V_b-V_a=C_3\left(\frac{1}{b}-\frac{1}{a}\right)$$
In 2D the integral is
$$V_b-V_a=-\int_a^b \frac{C_2dx}{x}=C_2(\log a-\log b)$$

So it would seem to follow that in 1D the field will be constant and so the difference in potential will be
$$V_b-V_a=-\int_a^b C_1\,dx=C_1(a-b)$$

Given two objects of the same charge, and no other charge sources between them, the field will be zero between them, hence the potential will be constant in that interval.