A fluoronium ion (H2F*) can be thought of as the superposition of 2 tiny electric dipoles that are oriented with an angle of 114.75° between them. The positive end of each dipole has a charge of qe and the negative end of each has (-qe/2), where qe = 1.6x1019 C. The separation distance d for each of the 2 dipoles is 9.5x1011 m.
A fluoronium ion (H2F*) can be thought of as the superposition of 2 tiny electric dipoles that are oriented with an angle of 114.75° between them. The positive end of each dipole has a charge of qe and the negative end of each has (-qe/2), where qe = 1.6x1019 C. The separation distance d for each of the 2 dipoles is 9.5x1011 m.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![A fluoronium ion (H2F*) can be thought of as the superposition of 2 tiny electric dipoles that are
oriented with an angle of 114.75° between them.
The positive end of each dipole has a charge of qe and the negative end of each has (-qe/2),
where qe = 1.6x1019 C. The separation distance d for each of the 2 dipoles is 9.5x1011 m.
95 pm
[Image source: Wikipedia]
114.75°
What happens to the fluoronium ion when it is placed in a uniform electric field (select all that
apply)?
It experiences a net torque about the midpoint of the fluoronium ion's resultant dipole vector
O It experiences a net torque about a point in between the midpoint and the tail the fluoronium ion's
resultant dipole vector
O It experiences no net force
O It experiences a net torque about a point in between the midpoint and the head the fluoronium ion's
resultant dipole vector
O It experiences no net torque
It experiences a net torque about the location of the fluorine (F) nucleus
O It experiences a net force in the direction parallel to the applied electric field](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe09d4462-f34d-4530-ac40-5ce52271c7b7%2F29eae71b-3f66-4472-a433-7921645a1aa2%2F89ysyh_processed.png&w=3840&q=75)
Transcribed Image Text:A fluoronium ion (H2F*) can be thought of as the superposition of 2 tiny electric dipoles that are
oriented with an angle of 114.75° between them.
The positive end of each dipole has a charge of qe and the negative end of each has (-qe/2),
where qe = 1.6x1019 C. The separation distance d for each of the 2 dipoles is 9.5x1011 m.
95 pm
[Image source: Wikipedia]
114.75°
What happens to the fluoronium ion when it is placed in a uniform electric field (select all that
apply)?
It experiences a net torque about the midpoint of the fluoronium ion's resultant dipole vector
O It experiences a net torque about a point in between the midpoint and the tail the fluoronium ion's
resultant dipole vector
O It experiences no net force
O It experiences a net torque about a point in between the midpoint and the head the fluoronium ion's
resultant dipole vector
O It experiences no net torque
It experiences a net torque about the location of the fluorine (F) nucleus
O It experiences a net force in the direction parallel to the applied electric field
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