projectile q (1 Viewer)

[GaganDeep]

can any one explain this using 3u approach lo.

 

[alcalder]

No real need to use the 3 Unit maths approach. You would only use 3 unit integration if they were going to ask you to do things like:

- find the distance the electron travelled,
- find the time it took for the electron to leave the electric field and so on.

a) The energy gained from the acceleration in the electron gun is transformed into Kinetic Energy.

E = qV
E = 1.6 x 10^-19 x 1000
= 1.6 x 10^-16 J = ½ mv²
Thus v = √ (2 x 1.6 x 10^-16/9.1 x ^-31)
v = 1.87 x 10⁷m/s

b) Now, the electron enters the Electric field of strength 10000 N/C.
The horizontal veolcity has no effect on this calculation. It would be like throwing a ball off a cliff with a certain velocity. The acceleration due to gravity still acts in a verticle direction unaffected by the horizontal motion.

F = ma = Eq
9.1 x ^-31 a = 10000 x 1.6 x 10^-19
Thus
a = 1.76 x 10¹⁵

 

[Naylyn]

It takes a lot longer using the integration method:
E=V/d
E=1000/d

also
F=qE
F=1000q/d

now since F=ma
ma=1000q/d
a=1000q/md
d(v^2/2)/dx = 1000q/md
v^2/2 = 1000qx/md + C
v^2 = 2000qx/md + C
now at x=0 v=0
0=0 + C
v^2 = 2000qx/md

now for the required velocity x=d
v^2 = 2000q/m

not really worth effort is it...

 

[sinist4]

dam.. for some reason that picture in the link is very small...

 

[The Gate Keeper]

sinist4... click the picture!

It gives you a nice BIG version of it.

 

[sinist4]

for some reason its still the same lol..ahh dw bout it
thanx dude anyway

 

[Naylyn]

Where does E=qV come from?

 

[helper]

Where does E=qV come from?

W=qV would have been better, so not confuse with E for electric Field strength

W=Fd
F=Eq
E=V/d

F=Vq/d
W=Vqd/d
W=qV

 

[Naylyn]

ah thanks, the unfamiliar character confused me lol