# Pd across the capacitors

• thereddevils
In summary, a parallel plate capacitor is charged to 20 V and then a simple pendulum with a metal bob is placed in between the plates. After 5 complete oscillations, 10% of the negative charge on the plate is transferred to the bob and neutralized when it touches the positive plate. This means that the potential difference between the plates has decreased by the same fraction. The same applies to the positive charge on the other plate. Therefore, the potential difference between the plates of the capacitor after 5 oscillations is 10% of the original 20 V, or 2 V.

## 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 .

## 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 .

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.

Stonebridge said:
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 ?

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.