Find the scalar value and direction of the electric field

  • Thread starter HJKL
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  • #1
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Homework Statement



Two charged balls are placed in point A and B and the distance between them is 9,54cm. Each of the balls are charged with 8,0 x 10^-8 C. Find the scalar value and direction of the electric field in point C placed 5 cm from A and 6 cm from B.

Homework Equations


Cosine Rule: a^2 = b^2 + c^2 - 2ab x cosθ
E = Ke x (q/r^2)


The Attempt at a Solution


[/B]
I have calculated the electric field of A and B.
EA = 287680 N/C
EB = 199778 N/C

After that I used cosine rule to calculate EC and got EC = 255341 N/C. This is correct.

But when I try finding the direction I calculate the angle between EC and EA (β) like this:
β = cos^-1 ((EA^2 + EB^2-EC^2)/(2EAEB)) = 60 degrees

The solution says that I should calculate the angle between EC and EA (β), but that the equation looks like this:
β = cos^-1 ((EC^2 + EA^2-EB^2)/(2ECEA)) = 42,65 degrees

Why am I supposed to use that equation?
 

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  • #2
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The wrong thing is your Cosine rule at the first line of your "relevant equation". Please check again :wink:
 
  • #3
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Oh! No, its not that. I just wrote it down wrong here:smile: I've used c^2 = a^2 + b^2 - 2ab * cosθ, and I get the first part right. I just don't get why I'm supposted to use this equation β = cos^-1 ((EC^2 + EA^2-EB^2)/(2EC EA)) to find the direction.
 
  • #4
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But...the answer in the solution is a correct answer, just compare to the Cosine rule :eek:
You wrote down the correct cosine rule, as above, but when you put ##E_{A}, E_{B}, E_{C}## into your calculation, I see that you put them in the wrong place
 
  • #5
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I was a bit confused about which angles goes with which side, but I get it now. Thank you!:smile:
 
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  • #6
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Hope this figure can help you :-p
 

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