# I Electric field strength with distance

1. Sep 24, 2016

### 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?

2. Sep 24, 2016

### 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.

Last edited: Sep 24, 2016