Electric Field & it's components

[victorializ]

Homework Statement

Two point charges, Q1 = -4.0 μC and Q2 = +3.0 μC, are placed as shown. In Figure 17.14b, the x-component of the electric field, at the origin O, is closest to:
A) -3000 N/C
B) 3800 N/C
C) 9800 N/C
D) -3800 N/C
E) -9800 N/C

Homework Equations

kq/r^2

The Attempt at a Solution

i understand how to get the magnitude of both forces of Q1 and Q2 by using kq/r^2
i found the hypotenuse to use as the length which is 2.57m .
what I'm lost on is the direction vectors that i need multiply the forces by.

in relation to Q2 i have found the inverse tangent and the angle comes out to be about 33. but when i multiply it by the force i get and get the difference with the force of Q1 it's not right.

any help is much appreciated, this one problem has been giving me trouble for quite some time!

 

[dauto]

Don't describe your calculation. Show it. If you don't show us your calculation how do expect us to find what's wrong with it?

 

[victorializ]

sorry here it is.
kq/r^2

Q1:
(9 x 10^9) x (-4 x 10^-6) / (2.3^2) = -6810 N/C

Q2:
(9 x 10^9) x (3 x 10^-6) / (2.75^2) = 3570.3 N/C
[the 2.75 distance I'm using is found - hypotenuse]

i'm unsure about the direction vector for Q1 since it is just on the x-axis ?
direction of vector for Q2:
atan(1.5/2/3) = 33.11 degrees
so 3570cos(33.11) = 2990 N/C

-6810 - 2990= -9800 N/C

the right answer is 3800 but I'm not sure why?

 

[victorializ]

Don't describe your calculation. Show it. If you don't show us your calculation how do expect us to find what's wrong with it?

sorry! I'm new to this!

 

[dauto]

sorry here it is.
kq/r^2

Q1:
(9 x 10^9) x (-4 x 10^-6) / (2.3^2) = -6810 N/C

Q2:
(9 x 10^9) x (3 x 10^-6) / (2.75^2) = 3570.3 N/C
[the 2.75 distance I'm using is found - hypotenuse]

i'm unsure about the direction vector for Q1 since it is just on the x-axis ?
direction of vector for Q2:
atan(1.5/2/3) = 33.11 degrees
so 3570cos(33.11) = 2990 N/C

-6810 - 2990= -9800 N/C

the right answer is 3800 but I'm not sure why?

why -6810 in the last line? why not + 6810? After all Q1 is negative isn't it?

 

[victorializ]

why -6810 in the last line? why not + 6810? After all Q1 is negative isn't it?

well with the calculations you get the -6810 . why would you change it; Q1 is to the right of the origin so wouldn't it just stay the same?

 

[haruspex]

but isn't Q1 positive since its to the right of the origin?

Q1 is negative, as given. The x displacement of the origin from Q1 is negative. The sign of the field comes from the product of the two, so is positive.
Just think about which way a positive test charge at the origin would be affected by Q1. It would be attracted to the right, yes?

 

[victorializ]

Q1 is negative, as given. The x displacement of the origin from Q1 is negative. The sign of the field comes from the product of the two, so is positive.
Just think about which way a positive test charge at the origin would be affected by Q1. It would be attracted to the right, yes?

oh yeah that makes much more sense, thank you!