Force between two point charges at given coordinates

[Pruddy]

Homework Statement

Determine the magnitude of the force between two point charges, Q1 = -2e located at (-3.60 cm, +4.10 cm) and Q2 = -8e located at (-0.60 cm, -8.60 cm). Give your answer in the form "a.bc x 10^(y) N".

Homework Equations

F = kq1q2/r2

The Attempt at a Solution

a.) I changed the unit vectors to meters, then I added, squared, and got the square roots of the vectors to get r = sqrt(((-0.036 +(- 0.006))^2)+((0.041+(-0.08))^2))
b.) Then I multiplied the value of "K" by the value of "8e" and "2e" and divided all three values by r^2

 

[Simon Bridge]

The Attempt at a Solution

a.) I changed the unit vectors to meters, then I added, squared, and got the square roots of the vectors to get r = sqrt(((-0.036 +(- 0.006))^2)+((0.041+(-0.08))^2))
b.) Then I multiplied the value of "K" by the value of "8e" and "2e" and divided all three values by r^2

Sounds reasonable - what was your question?
[edit] see ehild below - it is not really all that clear what your reasoning is.

Converting everything to SI units first does help make sure you get SI units out. However, you don't have to.

Where q_i=n_ie (n is an integer and e is the elementary charge) then |F_1,2| = (ke²)n₁n₂/r²

Since: ke² = 2.307x10^-24N.cm² you can do everything in cm.

 

[ehild]

Homework Statement

Determine the magnitude of the force between two point charges, Q1 = -2e located at (-3.60 cm, +4.10 cm) and Q2 = -8e located at (-0.60 cm, -8.60 cm). Give your answer in the form "a.bc x 10^(y) N".

Homework Equations

F = kq1q2/r2

The Attempt at a Solution

a.) I changed the unit vectors to meters, then I added, squared, and got the square roots of the vectors to get r = sqrt(((-0.036 +(- 0.006))^2)+((0.041+(-0.08))^2))
b.) Then I multiplied the value of "K" by the value of "8e" and "2e" and divided all three values by r^2

Why did you add the two vectors? The Coulomb force is inversely proportional to the square of the distance between two point charges. How do you get the distance between two points from their location? ehild

 

[hms.tech]

Treat the two given points as two position vectors in 2D.

Find the magnitude of the displacement between these two position vectors !

That gives you "r"

 

[Pruddy]

Hi, hms.tech
Thanks for your reply. I solved the two positions as two vectors using pythogoras theory. For example sqrt(((-0.036 +(- 0.006))^2)+((0.041+(-0.08))^2)), I do not know if it gives this solution gives me the distance between two point charges.

 

[Simon Bridge]

If you don't know if the method gives the answer you want, then why did you use it?

In general, if you have points $P=(p_x,p_y)$ and $Q=(q_x,q_y)$ defined against an origin $O=(0,0)$, then the distance between them is the length of the vector from P to Q which is given by change in position

$$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = (q_x-p_x,q_y-p_y)$$... that length would be: $$|\overrightarrow{PQ}| = \sqrt{(q_x-p_x)^2 + (q_y-p_y)^2}$$... your working does not look like this.

 

[Pruddy]

Hi,
I have tried solving this problem several times but my answers are still wrong.I don't know if the error is from the radius or if i am supposed multiply each charge by 1.602 x 10^(-19), since e = 1.602 x 10^(-19)

Attemped effort:
8.99 x 10^(9) x 2(1.602 x 10^(-19)) x 8(1.602 x 10^(-19))/.13050^2

= 2.17 x 10^(-26)


I have solved this problem a billion times but its still wrong. Please someone help. I will really appreciate it...:happy:)

 

[ehild]

Hi,
I have tried solving this problem several times but my answers are still wrong.I don't know if the error is from the radius or if i am supposed multiply each charge by 1.602 x 10^(-19), since e = 1.602 x 10^(-19)

Attemped effort:
8.99 x 10^(9) x 2(1.602 x 10^(-19)) x 8(1.602 x 10^(-19))/.13050^2

= 2.17 x 10^(-26)


I have solved this problem a billion times but its still wrong. Please someone help. I will really appreciate it...:happy:)

The magnitude is wrong. Try again.

ehild

 

[Pruddy]

Alright. But this what I got for the distance (r)

r^2 = (-0.006m-(-0.036m))^2 + (-0.086m-0.041m)^2
=.01703
square root of .01703 = .13050

 

[Muphrid]

Just put it in your calculator again and make sure you don't mistype and that all your order of operations is correct. You're off by some power of 10, which usually means you just didn't enter the calculation correctly.

 

[Pruddy]

The magnitude is wrong. Try again.

ehild

I have resolved it and my answer is 2.17 x 10^(-25). But the answer is still wrong...

 

[ehild]

I have resolved it and my answer is 2.17 x 10^(-25). But the answer is still wrong...

If you calculate accurately, it is 2.16 x 10^(-25) N. Have you added the unit (N)?


ehild

 

[Simon Bridge]

I have resolved it and my answer is 2.17 x 10^(-25). But the answer is still wrong...

I get 2.1676x10^-25N ... how many sig fig should you keep?
What makes you think this is the wrong answer?

 

[Pruddy]

I get 2.1676x10^-25N ... how many sig fig should you keep?
What makes you think this is the wrong answer?

Yes your right, the answer is correct... I was having a confusion with the units... Thank you

 

[Simon Bridge]

a.bc x 10^(y) N

ah riight. easy mistake ;)
One of the disciplines is to keep the units with the numbers when you substitute them into the equation. You get to use the laws of algebra on units too.

It's all good though.