Determine electric field at point P, Electric fields question

[riken9]

Homework Statement

Heres the probelm: http://imgur.com/TbzJxVa

Homework Equations

e = kq/r^2

The Attempt at a Solution

q-p
Ex = (9*10^9)(4)/(0.80)^2 = 5.625*10^10

Q-p
E1 = (9*10^9)(6)/(1)^2 = 5.4*10^10

Ex = E1sin45 = -3.82*10^10
Ey = E1cos45 = 3.82*10^10

P
p = SQRT((5.625*10^10)+(-3.82*10^10))^2+(3.82*10^10)^2) = 4.42 *10^10

The answer is actually 35 N/C, i don't know what am doing wrong. Help please!

 

[climb515c]

Looks at your charges, 4 and 6. They should be 4*10^-9 and 6*10^-9. Also double check your angle.

 

[riken9]

Why should they be *10^-9, sorry i was taught very poorly how to do these questions so i have limited knowledge on this topic.

 

[climb515c]

In the problem it states the charges are -6nC and 4nC. The "n" in nC is nano, which means 10^-9.

 

[riken9]

I found the angle to be 37 degrees, now my answer is around 49 Nc; which is still not close 35 Nc. Any more tips of what might of gone wrong?

 

[climb515c]

The angle is right.

It must be a small error in the algebra somewhere. If you keep getting it wrong, post your work here.

 

[riken9]

Heres what i am doing:

q-p
Ex = (9*10^9)(4*10^-9)/(0.80)^2 = 56.25

Q-p
E1 = (9*10^9)(6*10^-9)/(1)^2 = 54

Ex = E1sin37 = 32.5
Ey = E1cos37 = 43.1

P
p = SQRT((56.25)+(-32.5))^2+(43.1)^2) = 49.22

The 32.5 is negative because am thinking the - charge on Q makes the electric field go in ward, so its positive in the y-axis and negative in the x axis.

 

[climb515c]

Your cos and sin are wrong, you have them mixed up. Cos is for stuff dealing with "x" and sine is for stuff dealing with "y" (for this problem).

try it as Ex = E1cos37, and Ey=E1sin37.

All your reasoning is correct, just a simple mistake.

 

[riken9]

Thank you so much! To think, the only reason my answers were wrong cause of my stupid mistakes. Might need to be be careful, and not rush these questions.