# Find the scalar value and direction of the electric field

• HJKL
In summary, the conversation discusses the calculation of the scalar value and direction of the electric field at point C, which is located 5 cm from point A and 6 cm from point B. The equation used to find the direction (β) is β = cos^-1 ((EC^2 + EA^2-EB^2)/(2ECEA)), and the calculated angle is 42.65 degrees. The correct use of the cosine rule leads to a correct answer for the direction of the electric field.

## 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?

#### Attachments

• tempsnip.png
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The wrong thing is your Cosine rule at the first line of your "relevant equation". Please check again Oh! No, its not that. I just wrote it down wrong here 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.

But...the answer in the solution is a correct answer, just compare to the Cosine rule 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

I was a bit confused about which angles goes with which side, but I get it now. Thank you! • Nguyen Son
Hope this figure can help you #### Attachments

• HJKL