Question

What is the magnitude of the acceleration of an electron when in a electric field of 7.0 x 10^4 N/C?

Answer 1

F = q*E

m*a = q*E

a = q*E/m

= 1.6*10^-19*7*10^4/9.1*10^-31

= 1.23*10^16 m/s^2

Answer 2

Force on electron = 7.0*10^4*1.69*10^-19 C = 1.183*10^-14 N

mass of electron = 9.1*10^-31 kg

So acc = 1.183*10^-14/9.1*10^-31 = 1.3*10^16 m=s2

Answer 3

E= 7 * 10^4 N/C
m= 9.109x10^-31 kg
q= 1.6x10^-19 C
a=?

F=ma
F=qE>>>>> 1.6e-19x 7 * 10^4= 11.2 * 10^-15

11.2e-15 is the F so you do this...

F=ma>>>> 11.2e-15=9.109e-31a..... solve for a.....

11.2e-15/9.109e-31= 1.23e16

1.23e16 m/s2 answers

Answer 4

mass of electron = 9.1 x10^-31 kg

charge of electron = 1.6 X 10^-19

accel= charge X field /mass

accel= 1.2307 X 10 ^ 16

Answer 5

We will consider next the case of an electron entering a uniform electroc field between two parallel plates (Figure 4). The potential difference between the plates is V and the plates are aligned along the x direction and the electron enters the field at right angles to the field lines:
The force on the electron is given by the equation:

F = eE = eV/d = ma

There fore,

a=(e*E)/m

a = 1.2295*10^-31  m/s

mr Electron paths

Answer 6

force= electric field*charge

F=e*E

F=1.6*10^-19*7*10^4

F=11.2*10^-15

now,

acceleration,a=F/m

a=(11.2*10^-15)/(9.1*10^-31)

a=1.23*10^16