Physics (1 Viewer)

[jaychouf4n]

An electron, accelerated from rest, reached a speed of 3x10^5 m/s in a distance of 0.25m under the influence of an electric field. Calculate the magnitude of the electric field strength.

Ok the answer is 6.4x10^-2 NC^-1

But i keep getting wrong answers.

v^2=u^2+2ar
so a=2x(3x10^5)^2

F=ma
EQ=ma
Ex-1.602x10^-19=9.109x10^-31x(3x10^5)^2x2

Im not sure what i have to change to get the correct answer =/

 

[12o9]

Correct me if I'm wrong, but I think you've calculated the acceleration wrong

v²=u²+2as
beucase the particle starts from rest u=0
v²=2as
a=v²/2s
Sub v=3 x 10⁵ and s=0.25
a= [(3 x 10^5)^2]/2(0.25)

and I think your units are wrong for mass =/.

I'm not so sure though cause I couldn't get the answer either.

 

[minijumbuk]

Actually guys, I forgot almost everything in year 11 xD

But... I sorta used a calculus rule to do this because I forgot the formulas used =P

So... a = d(v² /2)/dx
So change in v² /2 = (3x10^5)^2 /2
Change in x = 0.25m

a = 1.8x10¹¹

F=ma
= 9.11x10^-31 (mass of electron) x 1.8x10¹¹
= 1.64x10^-19
F=qE
Therefore 1.64x10^-19=1.6x10^-19 (charge of electron) x E
E=1.02 N/C

I know it's not the answer you gave, but I couldn't find anything wrong with my solutions =\

 

[Aerath]

Actually guys, I forgot almost everything in year 11 xD

But... I sorta used a calculus rule to do this because I forgot the formulas used =P

So... a = d(v² /2)/dx
So change in v² /2 = (3x10^5)^2 /2
Change in x = 0.25m

a = 1.8x10¹¹

F=ma
= 9.11x10^-31 (mass of electron) x 1.8x10¹¹
= 1.64x10^-19
F=qE
Therefore 1.64x10^-19=1.6x10^-19 (charge of electron) x E
E=1.02 N/C

I know it's not the answer you gave, but I couldn't find anything wrong with my solutions =\

I would've gotten this too. But I would've gotten acceleration using 12o9's way.

 

[12o9]

Actually guys, I forgot almost everything in year 11 xD

But... I sorta used a calculus rule to do this because I forgot the formulas used =P

So... a = d(v² /2)/dx
So change in v² /2 = (3x10^5)^2 /2
Change in x = 0.25m

a = 1.8x10¹¹

F=ma
= 9.11x10^-31 (mass of electron) x 1.8x10¹¹
= 1.64x10^-19
F=qE
Therefore 1.64x10^-19=1.6x10^-19 (charge of electron) x E
E=1.02 N/C

I know it's not the answer you gave, but I couldn't find anything wrong with my solutions =\

Wouldn't you have had to change the units to grams =/?