I did these problems already, I am just unsure of my answers. As the answers to

Question

I did these problems already, I am just unsure of my answers.

As the answers to this problem involve vector calculations, be sure that your answers are complete and specify both magnitudes and vector components and directions. Assume SI Units. 30 points total

a) An alpha particle (helium nucleus, 2 neutrons + 2 protons) moves at a velocity of 1.0 x 102 i. What is the force experienced by the alpha particle if the value of the electric field is 5.0 x 10-2 j? What acceleration does it experience upon entering the electric field?

b) An electron moves with a velocity of 20.0 j and is subjected to a magnetic field of 20.0 x 10-3 k. What
is the magnetic force that acts on the electron? What would be the net magnetic force if a second magnetic field of 10.0 x 10-3 i were simultaneously applied?
What would be the net force if the second field were an electrical field of 20.0 x 10-2 j?

c) A proton moving at a velocity of 10.0 k enters a region of space containing all the fields as described in b). What force and acceleration does it experience?

A sphere with a charge of 2.0 x 10-6 C is placed a third of the distance between two negatively charged spheres, as measured from the left. The sphere to the left has a charge of -3.0 x 10-6 C and the sphere to the right has a charge of -6.0 x 10-6 C. A distance of 10.0 cm separates the end spheres. 30 points total

c) If the middle charge were a negative charge of 2.0 x 10-6 C, what would be the net force and net electrical field it would experience?

Explanation / Answer

(a) Force=q(vect(E)) So, Force = ( 2 X 1.6 X 10-19 X 5X10-2 j) (multiply with 2 for2 protons) =16 x 10-21 j So,vect( Accelaration)= vect(Force)/Mass So Vect(a)=(5 X 106)j (b) Vect( Magnetic Force) = q( vect(E) + (vect(v) X vect(B)) So,In case 1; Ans= -6.4 X 10-20 i N Note(negative charge of electron should be taken into account). So, in Case 2...Ans= (-3.2X10-20j-6.4x10-20i) N So, Magnitude = 10.32 X 10-20 N (c) Use Coloumb Force Formulae. Net Force= (21.6i) N Net Electric Field= (vect(F)/q)=-10.58 X 106 i (q=-2 X 10-6) In case of positive charge in the 1/3rd distance; Net Force= -21.6i N Net Electric Field= 10.58 X 106 i

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.