Trouble with electric force question

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
con31773
Messages
6
Reaction score
0

Homework Statement


I'm having a little trouble with an electric force question I have been working on. It seems simple, but my answer keeps coming out incorrect. Using the graph, deduce the electric force vector on charge q3 by q1/2. In case it is not clear, q3 lies on point (2,2)
Graph.jpg


Homework Equations



F=(kq1q2)/r^2*[itex]\hat{r}[/itex]

The Attempt at a Solution


Okay, well since electric force vectors simply add to give resultant force I can deduced the individual vectors and sum, as such.

F(1/3)=(kq1q3)/r^2*[itex]\hat{r}[/itex](1/3)

and the same applies to the second charge

F(2/3)=(kq3q2)/r^2*[itex]\hat{r}[/itex](2/3)

then I deduced r for both. Simply being 2 for q(2/3), and 5 for q(1/3) (from pythagoras)
Then, the unit vectors, as (2[itex]\hat{i}[/itex]+[itex]\hat{j}[/itex])/5 for q(1/3)
and [itex]\hat{j}[/itex] for q(2/3)

Put it together I got
F(1/3)=((kq1q3)/25)(2[itex]\hat{i}[/itex]+[itex]\hat{j}[/itex])

F(2/3)=((kq2q3)/4)[itex]\hat{j}[/itex]

Evaluate and sum, I got, 0.367[itex]\hat{i}[/itex]-0.412[itex]\hat{j}[/itex]

which of course is not correct. I'm sure there is a mistake in there but can not find it. Would be greatful for some fresh eyes to tell me where.

Thank you in advance. :)
 
Physics news on Phys.org
Hello con31772 and welcome to PF!

con31773 said:
then I deduced r for both. Simply being 2 for q(2/3), and 5 for q(1/3)

Oops. Check what you got for r for q(1/3).
 
Oops. Check what you got for r for q(1/3)

Haha, thanks very much. Could not find my mistake at all.