Physics Problem!!..Help!!

[suf7]

Can sum1 pleez help me with this problem???...i really dont have a clue???...i dont unsterstand what is the correct method to use???...need help!!

Three point charges are situated on the x-axis of a co-ordinate system.
Q1 = +2.0 nC is +0.05m from the origin
Q2 = -3.0 nC is + 0.1m from the origin
Q3 = -10.0 nC is -0.1m from the origin
What is the total force exerted by the three charges on a fourth charge “Q4 = +5.0 nC is situated at the origin??.....1 nC = 10-9C

[arildno]

Forces are additive.
So you must find out:
1) What is the magnitude and direction of the force from Q1 on Q4?
2) What is the magnitude and direction of the force from Q2 on Q4?
3) What is the magnitude and direction of the force from Q3 on Q4?
4) Then add the results from 1),2),3) together
5) Remember:
Whatever forces Q1,Q2,Q3 imparts ON EACH OTHER has NO RELEVANCE to the forces they impart on Q4!!!

[Nylex]

Use Coulomb's law to find the force acting on Q4 exerted by each of the other charges. The total force exerted on Q4 is given by the sum of those 3 forces.

[suf7]

Sorry but im really lost???...could you show me how to work out the first part???
What is the magnitude and direction of the force from Q1 on Q4?

ive never been shown any method for working these questions out, my teacher jus gives us the answers???...i just need to know what steps to take and how to achieve an answer????

[arildno]

State Coulomb's law, please.

[suf7]

force between two stationary point charges???

[arildno]

Precisely!

[suf7]

i know thats the law, i just dont know how to apply it???

[arildno]

Okay then, let's start at my 1)

Set up Coulomb's law to calculate the force from Q1 on Q4!

[suf7]

but i dont know how to???..i been given some formulas but i dont know which is which and which1 to use??..im confused??..sorry!!

[arildno]

Can you type in Coulomb's law, using Q1 and Q4 as your point charges?

[drnikitin]

Find forces (magnitude and direction) from each of the three charges on the fourth charge from Coulomb's law. Then add these forces taking into account their directions

[suf7]

is it this??

F = (Q1)(Q4)/4∏eor2

[suf7]

by the way how does nC come into all this???

[Nylex]

by the way how does nC come into all this???

nC = nano Coulombs. Charge is measured in Coulombs, so you need to convert your values into Coulombs (ie. multiply by 10^-9). The equation you have above for Coulomb's law was correct (assuming you mean r^2 on the bottom line).

[arildno]

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}}{(0.0 5m)^{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?

[suf7]

So is this right for the first point??

F = (Q1)(Q4)/4∏eor2
= (2*5) / (4∏)*(8.854*10^-12)*(2.5*10^-21)
= 3.597*10^31

the answer seems rather large???

[suf7]

thanks for that explanation..sorry, im abit slow..lol..
But wats hapend to the 4pi??and Eo??..wat have i done wrong in my calculation???

[Nylex]

So is this right for the first point??

F = (Q1)(Q4)/4∏eor2
= (2*5) / (4∏)*(8.854*10^-12)*(2.5*10^-21)
= 3.597*10^31

the answer seems rather large???

Why are you using 2.5 x 10^-21 as the distance between the charges? Also remember to convert your charges into coulombs if you want your force in newtons.

[arildno]

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.

[suf7]

i thought r = 0.05nC for Q1 to Q4 so i did (0.05*10^-9)
thats how i got (2.5*10^-21)??..i dont know wat to do??
how is it wrong??..help!!

[Nylex]

i thought r = 0.05nC for Q1 to Q4 so i did (0.05*10^-9)
thats how i got (2.5*10^-21)??..i dont know wat to do??
how is it wrong??..help!!

r is the distance between the charges, in metres.

[suf7]

i dont understand where you got the 10^-18 from on the top line???...and where does the 2.5m^2 and 10^-3 come from???..i just tried to follow the formula..lol..but i dint do it very well..lol

[Nylex]

i dont understand where you got the 10^-18 from on the top line???

He converted the charges into coulombs, from nano-coulombs. 1 nC = 10^-9 C and since q1 x q2 = C^2, you get (10^-9)^2 = 10^-18.

[arildno]

i dont understand where you got the 10^-18 from on the top line???...and where does the 2.5m^2 and 10^-3 come from???..i just tried to follow the formula..lol..but i dint do it very well..lol

Q1=2.0*10^{-9}C
Q4=5.0*10^{-9}C
Q1*Q4=10.0*10^{-18}C^{2}

Is that clear?

[suf7]

He converted the charges into coulombs, from nano-coulombs. 1 nC = 10^-9 C and since q1 x q2 = C^2, you get (10^-9)^2 = 10^-18.

y is this even on the top line for??? i thought its just q1*q2 on the top line???

[suf7]

Q1=2.0*10^{-9}C
Q4=5.0*10^{-9}C
Q1*Q4=10.0*10^{-18}C^{2}

Is that clear?

oh right..thanks..lol..i understand the top line,that makes complete sense now.... just the bottom line doesnt??

[arildno]

y is this even on the top line for??? i thought its just q1*q2 on the top line???
Yes, and
Q1=2.0*10^{-9}C
Q4=5.0*10^{-9}C
When substituting these values for Q1 and Q4, EVERYTHING on the right-hand sides of these equations must be included in the charges' places in the formula.
That's what I've done; you have not.

[arildno]

All right, you get the top line.
Can you now go back to post 20 and pinpoint what you don't understand?

[suf7]

Yes, and
Q1=2.0*10^{-9}C
Q4=5.0*10^{-9}C
When substituting these values for Q1 and Q4, EVERYTHING on the right-hand sides of these equations must be included in the charges' places in the formula.
That's what I've done; you have not.

nice1, i got that bit now, i dont think i converted into C, i left it as nC..thanks for clearing that for me.

[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 dont 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..:smile:
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 alot 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}}{(0.0 5m)^{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 cant thank you enough...i really appreciate all the time and effort you have spent on me, ive learnt alot...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, im grateful!!