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)
`3.0\cdot 10^(12)`

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

Step-by-step explanation:

a)

The charge on one electron is negative and equal to the fundamental charge, so the charge on one electron is

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

Protons instead are positively charged particles, so they have a charge of

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

This means that the net charge of an object can be written as follows:

`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

This can also be rewritten as

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

This equation gives the difference between the number of protons and the number of electrons in an object.

In this problem, object A has a charge of:

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

Substituting, we find

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

So, before the two objects come in contact, object A has
`3.0\cdot 10^(12)` electrons more than protons.

b)

When the two objects are touched together, the number of electrons transferred from object A to object B is:

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

So the total charge transferred from object A to object is

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

At the beginning, object B has a charge of:

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

So, the final charge on object B after the electrons are transferred will be:

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