An electron is released in a uniform electric field given by E = 4 k where “k” i

Question

An electron is released in a uniform electric field given by E = 4k where “k” is the usual unit vector and E is in N/C. What is the direction of (a) the force on the electron, (b) the acceleration of the electron, and (c) the rate of change of the momentum of the electron?

is E = 4k = E = 4(9x10^9)? or 3.6 x10^9

if thats so im getting

a)F = -5.76 x 10^(-9)

b)a = qE/m = -5.76 x 10^(-9) / 9.11 x 10^(-31) =   -6.32 x10^21

c)-5.76x10^(-9)

All directions are -x or -i.

another post i saw had these anwsers

F= -6.4x10-19 C K direction = -K
a= -70.329x106 m/s2 K direction = -K
Rate of change of momentum = -6.4x10-19 C K direction = -K

Please explain anwser throughly.

Explanation / Answer

Part A

E = 4k

k is a unit vector which only show that electric field is in z direction. its magnitude remain same.

E = 4N/C in +z direction

Part A

now force on electron (negative charge)is in opposite to the electric field so direction of force is negative z direction(-k)

F = Q*E

F =1.6*10^-19 * 4

F = 6.4*10^-19N (-k)

Part B

acceleration

a = F/m

m = 9.1*10^-31kg

a =  6.4*10^-19/9.1*10^-31

a = 0.703*10^12 m/s^2 in negative z direction

a = 0.703*10^12m/s^2 (-k)

Part C

rate of change of momentum = force = 6.4*10^-19 N (-k)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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