What will happen when X-rays are passed through a non-evacuated GLE? My teacher said that “The leaves would collapse due to ionization of air”. What does this mean?
[Physics] X-rays through a Gold Leaf Electroscope
electrostatics
Related Solutions
From gauss law we have $$\Phi_E=\frac{q_i}{\epsilon_0}$$ $q_i$ represents total change inside a closed surface and is independent of surface area and radius of that closed surface.
to the other end remember $$\Phi_E=\oint E.dA$$ where $$|E|=\frac{q}{4\pi\epsilon_0r^2}.$$
$\Phi_E$ is of course constant in your example (all questions including Q.1.)
and the reason the electric flux changed in question 1 is you are misunderstood.
Field will increase if $r$ decrease because $E\propto \frac{1}{r^2}$.
Q.3's answer : (taken freom H.C verma concepts of physics chapter 30 exercise 5): Since the charge is placed at the centre of the cube. Hence the flux passing through each side = $q/6\epsilon_0$
I have a guess, although I don't know if it is correct.
If you model the globe as a simple insulating circular glass shell with a constant spatial charge density embedded in the glass in the immediate vicinity of the finger and solve Maxwell's equations numerically for the potential, you observe something like this:
The rationale for placing a nontrivial static charge density in the vicinity of the finger in contact with the spinning globe is simply because glass, the so-called "vitreous" static source, dislodges surface charges during mechanical abrasion, and the charges have nowhere to go, since glass is mostly nonconducting.
As you can see, on the inside and outside of the glass in the immediate vicinity of the finger, there is a potential gradient. As such, dielectric breakdown becomes a possibility. By Paschen's Law, dielectric breakdown is orders of magnitude more likely in the partially-evacuated interior of the globe; as such, surface charge in the vicinity of the finger on the interior of the globe are accelerated through the vacuum and redistribute themselves on the lower-potential walls of the globe farther away from the finger. The electron collisions with nitrogen and other rarefied molecules in the vicinity of this transient current generate the blue glow near the finger.
Since the globe is spinning and the finger continually moves across the globe, there is no need to worry about a net nonphysical change in the interior surface charge density during each traversal of the loop, since they are constantly being "spatially recycled" during the spinning cycle.
For this reason, I am willing to bet that a similar experiment conducted with a finger rubbing the same spot on an evacuated glass bulb will NOT produce a continual glow discharge in the vicinity of the finger, despite the mechanical similarity of the processes.
Finally, assuming my mechanism is correct, there should be no visible glow discharge if the glass bulb is near-perfectly evacuated. Vacuum techniques in Hauksbee's time were, of course, far from this degree of rarefaction.
Best Answer
X-rays 'knock' one or more electrons from some of the air molecules, so the air is now partly ionised: there are positively charged ions and electrons (which may attach themselves to neutral molecules to make negative ions). The electrons (or negatively charged ions) are attracted to the leaf-and-plate assembly (if this had been positively charged) and surrender their excess electron(s), so gradually discharging the leaf and plate. The positive ions are attracted to the metal parts of the case of the electroscope, for these will have been charged negatively.