Three Point Charges in a Row, Find Magnitude 1.5cm left of middle charge

[fal01]

Homework Statement

Consider three charges arranged as shown.

+ + -
1 2 3
What is the magnitude of the electric field
strength at a point 1.5 cm to the left of the
middle charge? The value of the Coulomb
constant is 8.98755 × 10^9N
m2/C2.
Answer in units of N/C

q1=8.2 µC= 8.2*10^-6 C
q2=4.5 µC = 4.5*10^-6 C
q3=−4 µC= -4*10^-6 C

r1,2=4.4 cm= .044m
r2,3= 4.6 cm=.046m

kC= 8.98755*10^9

Homework Equations

F=kC*q1q2/r^2

The Attempt at a Solution

so I am guessing I need to start off with something like this...

F2,1=kC*q2q1/r2,3^s

F2,3= kC * q2q1/r2,3^2

and then find Fnet= F2,1 +F2,3

One of the main things I am struggling with is the 1.5cm (.015m) left of the center point.

 

[davo789]

You said in the original question that you wanted to find the electric field, but then started trying to calculate the force! I think the reason you're struggling is because you can't calculate the force of charges on a bit of chargeless space. Your logic is fairly sound otherwise!

 

[fal01]

Oh lol...

Then I use E=F/q=kq/r^2

then do I calculate the E for each one?
like so..


E1=k*q1/r^2
E2=k*q2/r^2
E3=k*q3/r^2

then what would r be?

 

[davo789]

What do you think r would be?

 

[fal01]

I am guessing the r should be different for each one...

E1=k*q1/r^2 r=.044m -.015m=.029m
E2=k*q2/r^2 r=.015m
E3=k*q3/r^2 r=.046m +.015m=.061m

 

[davo789]

Yes, exactly that!

 

[fal01]

So now
E1=8.76*10^7
E2=1.8*10^8
E3=9.66*10^6

so, Enet=E1-E2+E3

Did I get the signs right?

 

[davo789]

Yes, good spot!

 

[fal01]

Thanks so much! I got it! :)