Pd across the capacitors

[thereddevils]

Homework Statement

A parallel plate capacitor is charged by a battery to 20 V and then disconnected . A simple pendulum consists of a metal bob which is suspended by a nylom thread is placed in the middle of the plates of the capacitor . Then the bob is oscillated towards the negative plate of the capacitor . Each time when the bob touches the negative plate , 10 % of the charge at the negative plate will be transferred to the bob and these charges are then neutralised when the bob touches the positive plate of the capacitor . Determine the potential difference between the plates of the capacitor after 5 complete oscillation of the bob .

Homework Equations

The Attempt at a Solution

so i found that 0.1111 of the negative charges have been neutralised and 0.8889 of the -ve charges is left and i am not sure what to do next .

 

[Stonebridge]

The pd on the capacitor depends on the charge. If it has lost a certain fraction of its charge it should have "lost" the same fraction of its pd.

 

[thereddevils]

The pd on the capacitor depends on the charge. If it has lost a certain fraction of its charge it should have "lost" the same fraction of its pd.

thanks , am i correct in finding the percentage of -ve charge lost from the plate ? When it says the +ve charge from the other plate neutralise the -ve charge , is the +ve charges also decreasing at the same rate as the -ve charges ?

 

[Stonebridge]

Yes. If you imagine a charged capacitor that has +10 coulomb on one plate (A) and -10 on the other (B), we say that the capacitor has been charged to 10 coulomb. (Not 20)
If you transfer 1 coulomb of negative charge from plate B to plate A, then plate B has lost 1 coulomb and now has -9, while plate A is now +9. (The one negative charge having neutralised one of the positives.)
The result is that the capacitor now has a charge of 9 coulombs. +9 on the one plate and -9 on the other.