Suppose we have an ebonite rod (insulator). This rod has a negative charge, and once it touches a neutral pith ball, charges are distributed amongst the pith ball. However, why doesn't the same happen if we have neutral paper rather than a pith ball? Even though the paper is an insulator, shouldn't the charge migrate from the rod to the insulator and stay on the point of contact?
[Physics] Why can an insulator send charges to a conductor via conduction, and not upon an insulator via conduction
conductorselectrostaticsinsulators
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My answers are:
(1a) If the metallic rod is negatively or positively charged (excess or deficiency of electrons) and you touch the plate of the electroscope, part of the negative (positive) charge will be transferred to the electroscope because upon touching the rod+electroscope form one conductor with the same potential and the electrons distribute on its surface. Note: if the plate of the electroscope is replaced by a conductive beaker (Faraday's ice pail) and you insert the charged rod into the beaker, the whole charge will be transferred. NB: Negative charge is due to an excess electrons, positive charge is a deficiency of electrons. In both cases the charge moves due to the movement of the mobile electrons only.
(2a) When you approach a neutral electroscope's plate with a negatively charged rod, a separation of charges occurs on the electroscope before electrical contact by electrostatic induction, i.e. positive charges of the electroscope move towards the approaching negatively charged rod. This, of course, happens because the negatively charged electrons move away. When you touch, these positive charges will be neutralized by some of the negative charges of the rod and in the end the negative charge of the rod distributes partly over the electroscope and partly over the rod.
It's unclear whether you have a thin shell insulator of radius R and total charge Q with uniform surface charge ($\sigma=\frac{Q}{4\pi R^2}$) or a solid sphere with uniform volume charge ($\rho=\frac{3Q}{4\pi R^3}$), but the argument is the same.
The net electric field outside the spherical shape, at distance $R_0>R$ from the center of the sphere, is the vector sum of tiny electric fields, $d\vec{E}$, due to each point charge on/in the sphere. That tiny charge depends on the radius, $dQ=\sigma( dA)$ or $dQ=\rho (dV)$, and the tiny fields are $$d\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2}\hat{r}$$ where $r$ is the distance from the small charge to the point where you are calculating the field (point of interest), and $\hat{r}$ is a unit vector pointing from the small charge toward the point of interest. For some of the tiny charges $r<R$ and for others $r\ge R$. We can calculate $r$ using the law of cosines to get $$r^2=R^2+R_0^2-2RR_0\cos(\theta),$$ where $\theta$ is the polar angle with the polar axis along the sphere-center-to-point-of-interest line.
When you add up all these small electric fields due to small charges, the vector components which aren't radial all sum to zero. Also, the $r$ values are changing but $R$ value in the $dQ$ isn't, so some tiny charges are more influential than others. That is, each dQ has a small contribution to the total because the total charge isn't all in one location.
That's why the spherically distributed charge looks to the outside world to be the same as a point charge.
Apocryphally, this same question, except regarding the influence of Earth's mass, was Newton's motivation behind inventing the summing technique which we know as calculus. Before him, the concept of infinitesimally small things was abhorrent to mainstream mathematicians (apologies to Liebnitz).
Best Answer
The charges should also migrate to a piece of paper upon touching the ebonite rod. But probably the paper is not as good an insulator as the pith ball, which is possibly also hanging on an insulating thread.