What is the magnitude of the electric field

AI Thread Summary
To find the magnitude of the electric field at the point (0.506 m, 0.506 m) due to two charges, q1 = +21.3 nC and q2 = +11.0 nC, first calculate the distances from the point to each charge. Use the formula E = k * q / r² to find the electric field contributions from both charges. After determining the electric field vectors, resolve them into their x and y components and sum these components separately. Finally, apply the Pythagorean theorem to find the magnitude of the total electric field at the specified point. The overall goal is to obtain the magnitude of the total electric field at (0.506 m, 0.506 m).
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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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How would you think to approach the problem?
 
21.3+11.0 /1 = .506/.506 ?
 
Based on what?
 
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.
 
theres no angle though, so it would be .506 sin____ and .506cos___ or is it 90 or 180 degrees
 
xswtxoj said:
theres 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.
 
.506 sq + .506 sq = 574.69 then sq root = .715,
 
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.
 
  • #10
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?
 
  • #11
We're looking at distances, not charges.
 
  • #12
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?
 
  • #13
They want you to find the distance from 1,0 to .506,.506 for the second charge. That triangle then is .494,.506.
 
  • #14
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?
 
  • #15
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.
 
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