Recent content by tellmesomething

well yes a) 0 b) ##\frac{\sigma_1 A}{2\epsilon}+ \frac{\sigma_2 A}{2\epsilon} +\frac{\sigma_3 A}{2\epsilon}## (A here being the area of the pillbox) c) ##\sigma_1 + \sigma_2 + \sigma_3=0## d) ##\sigma_1= \frac{-3Q}{A}## (A here being the area of the infinite plates)

Like since at infinity potential is assumed to be 0, when we move from infinity (on either side) to plates A or D no work is done as these surfaces have 0 charge and there's no field to work against. When we move from plate A to C the potential difference also comes out to be 0 which is...

Yes but the field is not zero to the immediate right of plate 3, if we move from plate 4 to plate 3 we are doing work against a field, and that means there's a non zero potential, but this shouldn't be the case... this is what i was trying to ask in the post as well...

Oh right sorry that is probably the case, i copied it in class and the teacher did not give a proper question statement so i paraphrased, but yes this is what is implied, the plate 3 is grounded after.

This is the total charge on the plates initially before the third plate gets earthed. We have to find the charge distribution on the 4 plates after the third plate is earthed. The field you're talking about is before its earthed.

So, we know that since the 3rd plate is grounded that means the potential from the right side of the third plate to infinity should be 0, i.e no work should be required to bring a test charge from infinity to the right side of plate 3. Similarly the left side of plate 3's potential should also...

I apologise, I should have been more careful

It was only when I read post #3 that this point was clear to me, i wasn't thinking this way earlier..

Yes this was the intended question , my instructor wrote it in class on the blackboard and then he just solved it, this was the diagram provided and the homework statement was all that was given. Nevertheless I should have added the assumptions which was never written down in the question but...

do you mean as in the weight component of block B along the inclined plane exceeds even the maximum static friction? this is before spring force starts developing due to block A ?...

But since we are increasing the mass gradually, aren't we sure that only at the maximum extension of the spring we will get a new spring force to check if it is sufficient enough to just move block B? wouldn't using kx=mg here give us more than the minimum mass?...

The spring is relaxed when the system is released from rest. Yes... Yes i do mean that, sorry for the wording. After reading post #3 i think i get it, the mass of block A increases gradually, so if we didn't get the amount of force required to move block B even after the spring goes till...

Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static...

I obviously still have a long way to go to become like yourself and others on this site, but thanks:angel:

this was possibly the dumbest question i asked on here, at that time i had just started learning physics and had very wrong intuition/assumptions lol, i mean i was literally calling the "restoring force" reaction force as if action and reaction forces act on the same body o0). i also thought...