- #1
spaghetti3451
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- 34
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?
##\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?