Finding Horizontal Speed of Electron in Electric Field

AI Thread Summary
To find the horizontal velocity of an electron entering an electric field at an angle, use the formula for horizontal velocity, which is the initial speed multiplied by the cosine of the angle (5 x 10^6 m/s * cos(θ)). The vertical component can be determined similarly using sine (5 x 10^6 m/s * sin(θ)). The discussion highlights the relationship between sine, cosine, and tangent, noting that tan(θ) = sin(θ)/cos(θ) can be useful for finding angles when both horizontal and vertical components are known. Understanding these trigonometric relationships is essential for solving problems in particle dynamics within electric fields. The conversation emphasizes the importance of mastering basic trigonometric formulas for physics applications.
Amahia11
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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...
 
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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:​
 


tiny-tim said:
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!)
 
Hi Amahia11! :smile:
Amahia11 said:
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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