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To find the HORIZONTAL speed of an electron entering an electric field at an angle?

  • Thread starter Amahia11
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  • #1
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Homework Statement



So if an electron enters an electric field (between two charged parallel plates) at an angle (i.e not parallel to the x axis) and knowing that the speed is, say, 5 x 10^6 m/s, how would you find the HORIZONTAL velocity of the electron?

you know that horizontaly it must be (5 x 10^6)cos(theta) and that vertically it must therefore be (5 x 10^6)sin(theta) but how would you find the horizontal velocity?

please help, very appreciated!! :)



Homework Equations





The Attempt at a Solution



I thought it might help knowing that sin(theta)/cos(theta) = tan(theta) ?
But no idea how to find the horizontal speed...
 

Answers and Replies

  • #2
tiny-tim
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Hi Amahia11 ! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)

I don't understand … what are the unkowns, and what is given? :confused:

If 5 x 106 and θ are both given, then it's just 5 x 106 cosθ. :smile:
 
  • #3
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Hi Amahia11 ! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)

I don't understand … what are the unkowns, and what is given? :confused:

If 5 x 106 and θ are both given, then it's just 5 x 106 cosθ. :smile:
hi tiny-tim! thanks for the tips! :D

Ok, basically my physics exam on particle dynamics in electric fields is next week and my teacher told us to remember that sinθ/cosθ=tanθ (something which we btw never used in class)

now i understand (you're right) that the horizontal component will be 5 x 196cosθ but my question is how could you use the equation "sinθ/cosθ=tanθ"? I'm just trying to figure out how that could be useful in anyway... any ideas?

(sorry my initial post was quite unclear and vague!)
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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Hi Amahia11! :smile:
Ok, basically my physics exam on particle dynamics in electric fields is next week and my teacher told us to remember that sinθ/cosθ=tanθ (something which we btw never used in class)

…but my question is how could you use the equation "sinθ/cosθ=tanθ"? I'm just trying to figure out how that could be useful in anyway... any ideas?
ahhh …

yes, you will find tan = sin/cos useful, not when you're given the angle, but when you want to find it …

for example, if you know that the horizontal component of a force is 13, and the vertical component is 10, then the magnitude will be √(100 + 169), and the angle from the horizontal will have tan = 10/13 :wink:

btw, you should at some stage be learning the basic trigonometric formulas:

cos = adj/hyp
sin = opp/hyp
tan = opp/adj
(that's adjacent, opposite, hypotneuse :wink:)​
 

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