Why Do Electroscope Leaves Separate Wider When a Balloon is Quickly Moved Away?

  • Thread starter Thread starter username
  • Start date Start date
  • Tags Tags
    Electroscope
AI Thread Summary
When a charged balloon is brought close to an electroscope, the leaves separate due to the induced charge. Upon quickly moving the balloon away, the leaves can separate even further for a brief moment because they attempt to reclaim the charge deposited on the bottle's interior. This phenomenon occurs as the leaves are attracted to the opposite charges on the sides of the bottle, causing them to momentarily spread wider. If the balloon is removed slowly, the charge on the bottle's interior would dissipate, reducing this effect. The interaction highlights the dynamics of charge transfer and electrostatic forces in an electroscope.
username
Messages
226
Reaction score
2
Ok, this is not really homework so maybe it needs to be moved some where else.

Made a basic electroscope out of a wine bottle a while ago, charged up a balloon brought it close to the end of the scope and the leaves separated as expected, but when I quickly moved the balloon away from the scope the leaves separated 5 times wider for a few seconds then fell back. Why is this, is it due to some sort of sudden charge backwash in the process of depolarisation (is that a term :/) ?

Thx
 
Physics news on Phys.org
Not a "backwash" definitely. I think the following is correct:

the leaves were together before the balloon appraoached. The balloon came close and the leaves separated...

at this point, did the leaves come close to, or touch the sides of the bottle?

If they did, then they deposited a charge on the inside surface which then repelled the leaves slightly. THis would "squeeze" the leaves between opposing forces on both sides. When the balloon is taken away, the leaves desire to take back the charges that were deposited on the sides of the bottle (opposites attract, so all separated charges attract). The leaves come close to, or touch, the sides and pick up the charges and now its neutral again.
 
Voila, thank's that one has been bugging me for ages.

EDIT: I guess if I did not remove the balloon quickly away the charge on the inside of the bottle would then have time to dissipate reducing the effect.
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Replies
2
Views
10K
Replies
6
Views
3K
Replies
12
Views
5K
Replies
1
Views
4K
Replies
6
Views
3K
Replies
19
Views
5K
Back
Top