Electric Field Problem (why don't I multiply by sin and cos?)

[blackbrawler]

Homework Statement

In the figure the three particles are fixed in place and have charges q1 = q2 = +e and q3 = 2e. Distance a = 6.00 µm. What is the magnitude of the net electric field at point P due to the particles?

Homework Equations

E = k*q/r^2

The Attempt at a Solution

I looked in the back of the book and it shows that I shouldn't be multiplying by (cos45 + sin45) but I don't understand why since Electric Field is a vector value

here is my work

q1 and q2 cancel out so i only calculate q3

F = 9*10^9 * (3.2/(6^2/2)) * 10^-13 * (cos45 + sin45)

simplifies to
F = 160 * sqrt 2 = 226.274N

The actual answer is just 160N so that means I shouldn't multiply by (cos45 + sin45) even though that is the angle.

Edit: picture should be viewable

 

[Maybe_Memorie]

The image isn't displaying.

 

[blackbrawler]

The image isn't displaying.

can you see it now?

 

[flatmaster]

It's important to know that r is the physical distance between the charge and the point where you want to calculate the field. You can't plug in vector components for r because it's a scaler. I think you assumed that r=6 um, but you actually don't have the value for r yet. You need to find the actual distance between q3 and P.

 

[blackbrawler]

It's important to know that r is the physical distance between the charge and the point where you want to calculate the field. You can't plug in vector components for r because it's a scaler. I think you assumed that r=6 um, but you actually don't have the value for r yet. You need to find the actual distance between q3 and P.

i kind of skipped some of the steps but I did calculate r as sqrt(a^2/2) to get r^2 = 6um^2/2

 

[flatmaster]

i kind of skipped some of the steps but I did calculate r as sqrt(a^2/2) to get r^2 = 6um^2/2

Double check that.

 

[blackbrawler]

Double check that.

a^2 = 2r^2

(a^2)/2 = r^2

sqrt((a^2)/2) = r

 

[jackarms]

Electric field is a vector, but the question indicates that it's only asking for the magnitude, so that's why you can disregard the direction.

 

[blackbrawler]

Electric field is a vector, but the question indicates that it's only asking for the magnitude, so that's why you can disregard the direction.

but isn't the magnitude effected by the direction or angle?

If I'm applying a force at a 45 degree angle the amount it moves is different than if I apply it at a 90 degree angle.

 

[jackarms]

Depends. For torque it matters because the equation involves a vector product, which necessarily invokes direction. For electric field magnitude only depends on charges and distance, which are both scalars. Angle affects the amount of magnitude in the x and y directions, but not the total magnitude.

 

[haruspex]

F = 9*10^9 * (3.2/(6^2/2)) * 10^-13 * (cos45 + sin45)

There is no logical basis for the term (cos45 + sin45). If the field is E (vector) it has X and Y components |E|cos45 and |E|sin45, but you cannot add those together. To combine perpendicular components of a vector to find its magnitude you use the root-sum-square formula, which, of course, produce the answer |E|.