Hi, Ive done the first part of the exercise and just need help with the followin

Question

Hi, Ive done the first part of the exercise and just need help with the following question:

QUESTION: Assuming Q = e (positive unit charge) and r = 5 A, calculate the force acting on a Xe atom and compare the result with that acting on another unit charge e. Find the ratio of the two forces and comment on the strength of the polarization force compared to the Coulomb force. What happens to the ratio if the distance r is doubled?

A point charge Q is situated a large distance r from a neutral atom of atomic polarizability . Draw a labeled diagram of the situation. Find the electric field Eo at the location of the atom due to the point charge Q. rind-the eleetrie-field-F predueed by the indueedeipele-ef-the Answers can be expressed in terms of some or all of the quantities: Q and

Explanation / Answer

Formula for force between two charges is F = Kq1q2/r^2

here

Force between e and Xe ( has 54 atoms)

so

F1 = between e and 54 e is = (9e9 * 1.6 e -19 * 54 *1.6 e-19)/(5e-10*5e-10)

F1 = 4.97 e -8 N

F2 = force between e and e is F2 (9e9 * 1.6 e -19 * 1.6 e-19)/(5 e-10*5e-10)

F2 = 9.27 e -10 N

Noe the Ratio of these two forces is F2/F1 = (4.97 e -28/(9.27 e-30) = 54

---------------

if Radius is doubled, Force reduces by 4 times for each case

but ratio remains same

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