Finding the energy field due to three positive point charges on a 4th negative charge

[MarieWynn]

Homework Statement

Three equal positive point charge of magnitude Q=7.00uC are located at three corners of a square of edge length d=4.8cm. A negative charge of -21.00uC is placed on the fourth corner.
A) At the position of the negative charge, what is the magnitude of the electric field due to the three positive charges?
B) What is the magnitude of the attractive force on the negative charge?

Homework Equations

Ea=kQ/d^2 (also applies to Ec)
Eb=sqrt (Ea^2 + Ec^2)
e=KQ/r^2
r=sqrt(2d)

The Attempt at a Solution

I haven't gotten to part B yet, but here is the work I have done for part A.
Ea=(8.99x10^9)(7x10^-6)/.048^2=2.7x10^7
Ec=2.7x10^7
Eb=sqrt(Ea^2+Ec^2)=3.86x10^7
The electric field at the negative point due to the corner charge: e=KQ/r^2
e=(8.99x10^9)(21x10^-6)/0.3098^2=1967053.444
Since e and Eb are parallel, to find the magnitude of the electric field you would need to add them together: e+Eb=4.1x10^7

I have no idea why this is wrong. I also tried breaking it down into vectors, but that also got a wrong answer as well. If that was the right line of thinking I can post the work for that, but it felt off to me. I am including the diagram I drew up a well. I only have one try left so any guidance to help me see what I am doing wrong would be greatly appreciated.

 

[tiny-tim]

Hi Marie!

(try using the X² icon just above the Reply box )

e=(8.99x10^9)(21x10^-6)/0.3098^2=1967053.444

No, read the question again, the charge is still 7

(and where does 0.3098 come from?)

 

[MarieWynn]

Heh, I see the x² button now. :) Thanks!

0.3098 is r, where r=sqrt(2d).

I had originally kept the charge the same for that part where e=(8.99x10⁹)(7x10^-6)/r² but the answer was still wrong. It gave me 6.6x10⁵ and when adding that to E_b, I got 3.9x10⁷ N/C, which is still incorrect.

 

[tiny-tim]

0.3098 is r, where r=sqrt(2d)

nooooooooooo!

r=sqrt(2)d (or, easier, r² = 2d²)​

start again, and post it here first so we can check it!

then get some sleep! :zzz:​

 

[MarieWynn]

So r²=2d²=2*0.048²=0.004608
e=KQ/r²=(8.99x10⁹)*(7x10^-6)/0.004608=1.4x10⁷

Then adding that to E_b would give us: E= 5.2x10⁷ N/C
Is that right? Which would bring us to part B, which would be:
F=Eq=(5.2x10⁷)(21x10^-6)= 1092 N

 

[tiny-tim]

So r²=2d²=2*0.048²=0.004608
e=KQ/r²=(8.99x10⁹)*(7x10^-6)/0.004608=1.4x10⁷

Then adding that to E_b would give us: E= 5.2x10⁷ N/C
Is that right? Which would bring us to part B, which would be:
F=Eq=(5.2x10⁷)(21x10^-6)= 1092 N

Yes, that all looks ok , except you should have at least three https://www.physicsforums.com/library.php?do=view_item&itemid=523" in all your intermediate steps, or you might have a rounding error …

you always need more sig figs in the intermediate steps than in the final result

(and the Newtons should presumably be to only 2 sig figs)

do it again to check ​

 

[MarieWynn]

Thank you so much! This was correct and you really helped me to understand what I did wrong. When I get this problem right on the test I will thank you again! :)