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Electric Fields

Hi, could anyone help me in 2bi, I just don't seem to understand the question, why would moving the switch effect the electric field, if anyone could explain I would really appreciate it, Thanks.

Here the link: http://filestore.aqa.org.uk/subjects/AQA-PHYA4-2-W-QP-JUN10.PDF


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Jimmy20002012
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Original post by uberteknik
The ball is attracted to the top plate which therefore has a net +ve charge.

If the switch is moved over, the charge stored on the top plate is then shared between both plates. Therefore the ball will be attracted to both plates with the same force. It will therefore return to the centre position again.


Okay, but how does that make the electric field strength come to zero?


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Original post by Jimmy20002012
Okay, but how does that make the electric field strength come to zero?


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If the charge on both plates is the same and both now +ve, the electric field between the plates must be zero.

However, the ball was already displaced (stored potential energy) at the time the charge is equalised. This means that the attraction of the ball to each plate is now imbalanced since F = kQ1Q2/d2


The spring provides a restoring force, so the ball will now accelerate towards the central position again and in so doing the stored potential energy is converted to kinetic.

Can you now see how SHM is set up?
Original post by Jimmy20002012

Sort for the hassle but could you help me in the questions below as well because I have no idea ow to calculate change of momentum?

Thanks soo much

ImageUploadedByStudent Room1392032286.438264.jpg


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Can you please start a new thread for a new question, especially when the question is not related to the original post.

Thanks.

PS. I've done it for you.
(edited 10 years ago)
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Jimmy20002012
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Original post by Stonebridge
Can you please start a new thread for a new question, especially when the question is not related to the original post.

Thanks.

PS. I've done it for you.


Stone bridge could you help with that momentum question please?? :smile:


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