Question

In an experiment Erin holds two objects. Object A has a net charge of –4.81 × 10–7 C, and object B has a net charge of 6.41 × 10–7 C. She touches the objects together.

(a) Roughly how many more electrons are there than protons in object A before the two objects come in contact?

(b) After Erin touches the objects together, about 5.00 × 1011 electrons transfer from object A to object B. What is the new net charge on object B?

Answer 1

a) object A has
`3.0\cdot 10^(12)` electrons more than protons

b)
`+5.61\cdot 10^(-7) C`

Step-by-step explanation:

a)

Electrons are negatively charged particles, so they have a charge of

`q_e =-e= -1.6\cdot 10^(-19)C`

On the other hand, protons are positively charged particles, so they have a charge of

`q_p =+e=+1.6\cdot 10^(-19)C`

Therefore, the net charge of an object can be written as

`Q=N_p q_p + N_e q_e=e(N_p-N_e)`

where

`N_p` is the number of protons

`N_e` is the number of electrons

So the equation can be rewritten as

`N_p - N_e = (Q)/(e)`

Which gives the difference between number of protons and number of electrons.

Here, object A has a charge of

`Q=-4.81\cdot 10^(-7)C`

Therefore,

`N_p-N_e=(-4.81\cdot 10^(-7))/(1.6\cdot 10^(-19))=-3.0\cdot 10^(12)`

Which means that object A has
`3.0\cdot 10^(12)` electrons more than protons.

b)

When Erin touches the objects together, the number of electrons transferred from object A to object B is

`N_e' = 5.0\cdot 10^(11)`

The charge corresponding to these electrons transferred is

`Q'=N_e' (-e)=(5.0\cdot 10^(11))(-1.6\cdot 10^(-19))=-8\cdot 10^(-8) C`

The initial charge on object B was

`Q_B=+6.41\cdot 10^(-7)C`

This means that the final charge on object B after the electrons are transferred will be:

`Q=Q_B+Q'=+6.41\cdot 10^(-7)+(-8\cdot 10^(-8))=+5.61\cdot 10^(-7) C`