Four Equidistant Particles in electric fields

[tiny-tim]

ok so I found that with the formula F=kq1q2/r^2 in the diagonal direction where the charges are being repulsed that F= 6.50731 E-15 and diagonal direction where charge are drawn inward F=-37.1554, if my calculations are correct, would I just take the dot product and multiply them together to get the magnitude?

Sorry, C6ZR1, but you seem clueless as to what a vector is.

You don't add two perpendicular vectors by taking their dot product.

Let's try this …

If I add a vector of 3 in the x direction to 4 in the y direction, what do i get?​

 

[C6ZR1]

well that would be 3i+4j, correct?

Would I do the cross product?

 

[SammyS]

well that would be 3i+4j, correct?

That's correct for adding the vectors tiny-tim asked about.

Would I do the cross product?

tiny-tim is telling you to add the Electric Field due to each charge, adding them as vectors. There is NO product involved in that operation. It's adding, not multiplying.

 

[C6ZR1]

ok I think I'm completely lost.

I find the electric field from each charge:

E1= (8.99E9*1.602E-19*7.59E-9) /(0.065)^2= 2.58724 E-15 C

E2= (8.99E9*1.602E-19*-1.09E-8) /(0.065)^2= -3.71554 E-15 C

E3= (8.99E9*1.602E-19*1.15E8) /(0.065)^2= 39.2007 C

E4= (8.99E9*1.602E-19*-6.06E-9) /(0.065)^2= -2.0657E-15 C


Would I resolve them into their components? ie. E1= cos(45)2.58724 E-15 C+sin(45)2.58724 E-15 C etc?

Sorry for the trouble, I don't know why I'm not comprehending this

 

[tiny-tim]

Hi C6ZR1!

(just got up :zzz: …)

well that would be 3i+4j, correct?

yes

Would I do the cross product?

why?

as SammyS says, 3i+4j is the answer

when you add 3i to 4j (as vectors), the answer is 3i+4j

(obviously, it's more complicated if the two vectors aren't perpendicular )

(alternatively, you can say that it's 5 in the direction whose tan is 4/3)

 

[C6ZR1]

the thing is for the answer it says

Number: ____ N/C or V/m

and I cannot use vector letters, that's what is confusing me

 

[tiny-tim]

yes, the question asks for the magnitude of the total field, which is of course a scalar (not a vector) …

you get this by adding the individual fields as vectors, so that gives you the a total vector, and then you write down the magnitude of that vector

(in our example, the total vector was 3i+4j, and so the magnitude was 5)

 

[C6ZR1]

ohhhhhhhhhhhh ok. This makes sense now. lol Thanks for your time and help. Physics is starting to make more sense now