Electric field from two charges at (0.506, 0.506)

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Homework Statement



Two electric charges, q1 = +21.3 nC and q2 = +11.0 nC, are located on the x-axis at x
= 0 m and x = 1.00 m, respectively. What is the magnitude of the electric field at the
point x = 0.506 m, y = 0.506 m?
 
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They want you to find the E-field vector at .506,.506.

First calculate the |E1| and |E2| based on the geometric distance to both charges.

Then separate those vectors - they are vectors - into their components.

Then add the x-components and y components separately.

Happily they only want the magnitude of the E-field - |E| - so just use Pythagoras to figure the |E| of the resulting vector.
 
there's no angle though, so it would be .506 sin____ and .506cos___ or is it 90 or 180 degrees
 
xswtxoj said:
there's no angle though, so it would be .506 sin____ and .506cos___ or is it 90 or 180 degrees

You have the coordinates of the point.

and you know - or should know that a2 + b2 = c2

So ... figure it out.
 
xswtxoj said:
.506 sq + .506 sq = 574.69 then sq root = .715,

It is .715 but your intermediate result is nonsense.

So that's the r for 1 charge.

Now figure the r for the other.
 
21.3 sq + 11 sq= 574.7 equals 23.97 after sq rt, then take the 2 r's and add them up and tan y/x?
 
would it be : E1= ke q1/ r sq? then E2= ke q2/r sq but how would the 2nd r be found would the x= 0, and 1 or 1, 0.506 since it for q2?
 
They want you to find the distance from 1,0 to .506,.506 for the second charge. That triangle then is .494,.506.
 
once i get r, i solve for e, then once i get both e's do i add them up or leave it as 2 separate answers?
 
xswtxoj said:
once i get r, i solve for e, then once i get both e's do i add them up or leave it as 2 separate answers?

Once you get the |E| (magnitude of E) then you must resolve them into their x,y components and add the components - being careful of the signs - and then determine the magnitude of the result.

They want the magnitude of the Total E field at that point.