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D) activation gate - opens during depolarization
These are my answer if you want to use them
Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "
Solution
The force exerted on a charge q by an electric field E is given by
</span>

<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span>

with

,

is the charge on the bead. We want to calculate the field at

:

The proton has a charge of

, therefore the force exerted on it is

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass,

, we have


The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
The magnetic force on a charged particle is given as: F = qvBsinФ
B is the magnetic field.
q is the charge on the particle.
v is the velocity of the charged particle.
Ф is the angle between magnetic field (B) & the velocity of charge (v).
⇒ F = qvb sin Ф
Here, Ф = 90 degree.
and sin 90 = 1
Putting the values in the equation, we get,
F = 1.6 x 10^-19 C × 4.5 × 10^7 m/s × 1.6 T
= 11.52 × 10^-12N
= 1.2 x 10^-11 N up