Physics Problem Help: Find Total Force on Q4

[suf7]

You should have (I include the sign (direction) for the force):
F=-\frac{10.0*10^{-18}C^{2}}{4\pi*\epsilon_{0}*2.5m^{2}*10^{-3}}
If I understand your calculation correctly, you placed the 10^{-18} factor in your denominator, rather than in your numerator.

its the "2.5m^2" and the "10^-3", i don't know where they come from??

 

[arildno]

All right!
The DISTANCE between Q1 and Q4 (that is, what I've called r_{41})
is:
r_{41}= 0.05m=5*10^{-2}m
Hence,
r_{41}^{2}=25*10^{-4}m^{2}=2.5*10^{-3}*m^{2}
is that cleared up now?

 

[suf7]

All right!
The DISTANCE between Q1 and Q4 (that is, what I've called r_{41})
is:
r_{41}= 0.05m=5*10^{-2}m
Hence,
r_{41}^{2}=25*10^{-4}m^{2}=2.5*10^{-3}*m^{2}
is that cleared up now?

ok, i think that makes sense...so is this right??
f = (10*10^-18) / (4pi)(Eo)(2.5*10^-3)
= 3.597*10^-5

 

[arildno]

That seems much better..
And in which direction does this force push Q4?
(That's VERY important to know!)

 

[suf7]

thanks..phew, uve actually taught me soemthing, thanks to you it makes a lot more sense..your explanations have been really good, sorry if i annoyed u with my lack of intelligence..lol..erm, not sure, how can u tell the direction of push??

 

[arildno]

Can you now go back to my post "16" and see if you understand my reasoning there?
(Don't let the vectors frighten you!)
Just post again if there's something you don't get there..

 

[suf7]

That is correct, as long as you remember that the implied direction is AWAY from Q4.
On vector form, we have:
\vec{F}_{41}=\frac{1}{4\pi\epsilon_{0}}\frac{Q_{1}Q_{4}}{r_{41}^{2}}\frac{\vec{r}_{4}-\vec{r}_{1}}{r_{41}}
EXPLANATIONS:
1)Here, I have included the last fraction to have the basic direction explicitly included.
2)\vec{F}_{41} means: The force acting on Q4 from Q1
3) r_{41} is the DISTANCE between Q1 and Q4, that is, some positive number.
CALCULATIONS:
Q1=2.0 nC
Q4=5.0 nC
Hence, Q1*Q4=10.0(nC)^{2}
\vec{r}_{4}=\vec{0} (that is, situated at the origin)
\vec{r}_{1}= 0.05m\vec{i}
(that is, situated 0.05m to the right-hand side of Q4 at the origin)
Hence, \vec{r}_{4}-\vec{r}_{1}=-0.05m\vec{i}
That is, the direction indicated by this quantity is leftwards.

r_{41}=0.05m that is the DISTANCE between Q1 and Q4 is 0.05m

Collecting all together, we have:
\vec{F}_{41}=-\frac{1}{4\pi\epsilon_{0}}\frac{10.0(nC)^{2}}{(0.05m)^{2}}\vec{i}

That is, the force on Q4 from Q1 is directed leftwards.
Q4 tends to be REPELLED from Q1, because they have EQUAL TYPE OF CHARGE!
Was this okay?

isnt my calculation incorrect because i never put the "-" in front of the 0.05??
so that makes the direction negative right?

 

[arildno]

isnt my calculation incorrect because i never put the "-" in front of the 0.05??
so that makes the direction negative right?

I chose not to pick on that, because I sensed there were more pressing issues at hand.
In addition, since your calculation basically (and after a while, correctly) gave how "big" the force should be, I thought I might postpone the introduction of the sign.
Note that I did include it in my post (20) as well..

You are right, the minus sign gained from (16) shows that the force on Q4 works in the negative direction.
This expresses the simple fact that since Q1 and Q4 are electrical charges with the SAME SIGN, they will tend to REPEL each other.

 

[suf7]

that is really great...i can't thank you enough...i really appreciate all the time and effort you have spent on me, I've learned a lot...i think i should be able to complete the question now and when i have finished it i will post it and "HopeFully" it will be correct...lol...thanks again, you are a very good teacher!

 

[arildno]

No prob!
As you probably can see, what remains now is to calculate the force from Q2 on Q4 and the force from Q3 on Q4, and add the results together.

Be careful with your signs!

 

[suf7]

Q1 - Q4: = -3.597*10^-5
Q2 - Q4: = 1.348*10^-5
Q3 - Q4: = -4.49*10^-5

(-3.597*10^-5)+(1.348*10^-5)+(-4.49*10^-5) = -6.74*10^-5


have i made any mistakes?...is this the right answer?...-6.74*10^-5

 

[arildno]

You have the right signs, and the magnitudes look all right.
So, on the whole (without having calculated the quantities myself), it definitely smells like a correct answer.

 

[suf7]

thanks for your help, I am grateful!