At the instant shown here, a singly-ionized ^4_2 He atom (He^+ion) is at the mid

Question

At the instant shown here, a singly-ionized ^4_2 He atom (He^+ion) is at the midpoint between two infinite parallel plates. Its velocity is 7500 m/s toward the top of the page. The plats are separated by 20. cm and have a potential difference of 4200 v. Determine the magnitude of the electric force acting on the He^+ ion. Show your work. The direction of this electric force on the ion is; to the right to the left into the page out of the page toward top of the page toward bottom of the page How would your answer to part (a)&(b) change if the He^* ion were replaced by... L an alpha - particle weaker force, same direction same force, same direction stronger force, same direction weaker force, opposite direction same force, opposite direction stronger force, opposite direction a proton weaker, same direction same force, same direction stronger force, same direction weaker force, opposite direction same force, opposite direction In order to allow the He^+ ion to continue upward on an undeflected straight- line path, we can impose a uniform magnetic field in the region between the plant. this B-field must point in which direction to the right to the left into the page out of the page toward top of the page toward bottom of the page Determine the magnitude of this magnetic field. show your work.

Explanation / Answer

A) magnitudce of electric force F = q*E

E is the electric field between the plates = V/d = 4200/0.2 = 21000 V

q is the charge of the ion = 1.6*10^-19 C

then F = 1.6*10^-19*21000 = 3.36*10^-15 N

B) to the left

C) alpha particle has more charge than the He+ so,it experiences a strongerforce ,same direction

so theanswer is C) stronger force,same direction

for a proton

same force and same direction

D) according to flemings left hand rule the direction of B should be out of the page

e) mangitude of magnetic filed is B

since the force should be balnced

q*E = q*v*B

B = E/v = 21000/4200 = 5 T

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