Question

An electron is traveling down with an initial speed of 4.2x10³m/s at time t=0.  There is a constant electric field in the area of 5.9 N/C up.

electron У 4. 2x103 m/s

A. What is the direction of the force on (and acceleration of) the electron?  Why?  (looking for a sentence relating the directions of E and F and the sign of the charge.)

B. Will this force speed up or slow down the electron?  Explain why (based on the direction of the force compared to the direction of the velocity.)

C. After the electron has traveled .073 mm, find the velocity of the electron.  [1.3x10⁴m/s]

Answer 1

1.

Direction of Force (and acceleration) on the electron is downward. Reason:

Relation between force and electric field is given by:

F = q*E

Since q = charge on electron (which is negative) = -1.6*10^-19 C

F = -e*E

Since Electric field is upward and charge in field is negative, So force on electron will be in opposite direction of electric field, which is downward.

2.

Now Using Force balance

Fnet = q*E

From newton's 2nd law:

Fnet = m*a

So,

m*a = q*E

a = q*E/m

Since acceleration is downward and initial velocity is also downward, So this force will speed up the electron.

3.

Now Using given values, acceleration will be

m = mass of electron = 9.1*10^-31 kg

a = -1.6*10^-19*5.9/(9.1*10^-31)

a = -1.037*10^12 m/sec^2

Vi = -4.2*10^3 m/sec

S = distance traveled downward = -0.073 mm = -0.073*10^-3 m

(Here all the negative signs only represents the direction)

Using 3rd kinematic equation:

Vf^2 = Vi^2 + 2*a*S

Vf = sqrt ((-4.2*10^3)^2 + 2*(-1.037*10^12)*(-0.073*10^-3))

Vf = 13001.6 m/sec

Vf = 1.3*10^4 m/sec

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