A proton at location A makes an electric field E vector_1 at location B. A diffe
ID: 250001 • Letter: A
Question
A proton at location A makes an electric field E vector_1 at location B. A different proton, placed at location B, experiences a force F vector_1. If |E vector_1| = 400 N/C, what is |F vector_1| |F vector_1| = N Now the proton at B is removed and replaced by a lithium nucleus, containing three protons and four neutrons. The proton at location A remains in place. What is the magnitude of the electric force on the lithium nucleus |F vector on Li| = Now the Lithium nucleus is removed, and an electron is placed at location B. The proton at location A remains in place. What is the magnitude of the electric force on the electron |F vector on e^-|indicates the direction of the force on the electron due to the electric fieldExplanation / Answer
we know that
E =F / q
F = Eq
Force on proton F1 = E1 q
= ( 400 N/C ) ( 1.6 *10-19 C )
= 640 *10-19 N
According to coloumbs law
F 1 = k q1 q2 /r2
640*10-19 N = ( 9*109 )( 1.6 *10-19 C )2 /r2
The distancebetween two protons
r = [ ( 9*109 ) ( 1.6*10-19 C )2 / 640 *10-19 N ]1/2
= 1.90*10-6 m
char on lithium q2 = 3 P =3 ( 1.6 *10-19 C)
The magnitude of the electric force on the lithium nucleus
F = k q1 q2 /r2
= ( 9*109 ) 3( 1.6*10-19 C )2 /(1.90*10-6 m )2
= 1.91 *10-16 N
magnitudeof the electric force on the electron
F = k q1 q2 /r2
= ( 9*109 ) ( 1.6*10-19 C )2 /(1.90*10-6 m )2
= 6.38 *10-17 N