Answer:
Electric force, [tex]F=8.16\times 10^{-15}\ N[/tex]
Explanation:
A protein molecule in an electrophoresis gel has a negative charge. Electric field, E = 1700 N/C
There are 30 excess electrons. The charge on 30 electrons is, q = 30e
[tex]q=30\times 1.6\times 10^{-19}\ C=4.8\times 10^{-18}\ C[/tex]
The electric force in terms of electric field is given by :
[tex]F=q\times E[/tex]
[tex]F=4.8\times 10^{-18}\ C\times 1700\ N/C[/tex]
[tex]F=8.16\times 10^{-15}\ N[/tex]
So, the magnitude of the electric force on a protein is [tex]8.16\times 10^{-15}\ N[/tex]. Hence, this is the required solution.
Answer:
8.16 x 10^-15 N
Explanation:
Number of excess electrons = 30
Electric field, E = 1700 N/C
Charge of one electron, e = 1.6 x 10^-19 C
Total charge, q = charge of one electron x number of electrons
q = 30 x 1.6 x 10^-19 = 48 x 10^-19 C
The relation between the electric field and the force is given by
F = q E
F = 48 x 10^-19 x 1700
F = 8.16 x 10^-15 N
