A) Electrons
B) From the sphere to the rod
C) [tex]3.75\cdot 10^{10}[/tex]
Explanation:
A)
There are two types of electric charges that can be transferred:
- Negative charge: this is typically transferred by electrons, which reside in the outer parts of the atom. Electrons are negatively charged, and they have a charge of [tex]-e=-1.6\cdot 10^{-19}C[/tex] (fundamental charge).
- Positive charge: this is carried by the protons, which actually reside in the nucleus of the atom. The charge of a proton is [tex]+e=+1.6\cdot 10^{-19}C[/tex].
Since protons are inside the nucleus of the atom, they usually don't move around, so in normal conditions the electric charge between different objects is carried by electrons, not protons.
Therefore in this case, the charged particles transferred between the rod and the sphere are electrons.
B)
In this problem, we have:
- At the beginning, the glass rod has a net positive charge of
[tex]q_i=+14.0 nC[/tex]
- After the process and having touched the metal sphere, the glass rod has net positive charge of
[tex]q_f=+8.0 nC[/tex]
This means that the glass rod in the process has lost a positive charge of
[tex]\Delta q=+14.0-8.0=+6.0 nC[/tex]
Since we said that the charge carriers are the electrons (not the protons), in order for the glass rod to lose a positive charge, it means that it has actually gaind a negative charge: therefore, this means that electrons have moved from the metal sphere to the glass rod.
C)
In part B, we calculated that the net charge gaind by the glass rod is
[tex]\Delta q=+6.0 nC[/tex]
Converting into SI units,
[tex]\Delta q=+6.0\cdot 10^{-9}C[/tex]
We said that the charge carriers in this case are the electrons: so, the actual charge transferred from the metal sphere to the rod is
[tex]\Delta q=-6.0\cdot 10^{-9}C[/tex]
This charge actually consists of N electrons, so we can write:
[tex]\Delta q =(-e)N[/tex]
where
N is the number of electrons
[tex]-e=-1.6\cdot 10^{-19}C[/tex] is the charge of each electron
And solving for N, we find:
[tex]N=\frac{\Delta q}{-e}=\frac{-6.0\cdot 10^{-9}}{-1.6\cdot 10^{-19}}=3.75\cdot 10^{10}[/tex]
