Coulomb's Law and charged particles

[UNG]

you shouldn't be multiplying by cos45 here... the force between 1 and 3 is vertical... so there is no horizontal component.

Oh yeah that's right.
It would be by 90 degrees because its horizontal,
which means cos 90 = 0 and sin 90 = 1

Thanks

 

[UNG]

Sorry I have completed the first question,
does this look about right?

X component = -25 x 10-3 i N

y component = 46.48 x 10-3 j N

Net force = 39.18 x10-3 i j N

Are the units right? Should have a dash above the top of i and j?

 

[learningphysics]

I'm not getting those...

for the y-component I get -46.5*10^-3

For the x-component it appears you got the x-component of the force of 2 on 3 (but which should be +25*10^-3 not -25*10^-3)... but forgot to add the force of 4 on 3 which is also in the x-direction...

 

[UNG]

I see so I got the same answers, but mixed up the signs you say?

Which means that the total componets added together (Net Force) is wrong and should be a minus answer.

Therefore

Net Force = - 21.5 x10^3

?

 

[learningphysics]

No, you're missing the force of 4 on 3. please show your calculations in detail... where does the force exerted by 4 on 3 come into your calculations?

 

[UNG]

Well from the calculations it would make sense to add in a third part into the formula.

Something like this ?


$$\sum{\vec{F}_{3_{x}}} = {\left(\frac{{k_{e}}{|q_{3}|}{|q_{1}|}}{{{r}_{31}}^{2}}\right)}{cos{\theta_{31}}}{\hat{i}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{2}|}}{{{r}_{32}}^{2}}\right)}{cos{\theta_{32}}}{\hat{i}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{4}|}}{{{r}_{34}^{2}}\right)}{cos{\theta_{34}}}{\hat{i}}$$

$$\sum{\vec{F}_{3_{y}}} = {\left(\frac{{k_{e}}{|q_{3}|}{|q_{1}|}}{{{r}_{31}}^{2}}\right)}{sin{\theta_{31}}}{\hat{j}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{2}|}}{{{r}_{32}}^{2}}\right)}{sin{\theta_{32}}}{\hat{j}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{4}|}}{{{r}_{34}^{2}}\right)} {sin{\theta_{34}}}{\hat{j}}$$

 

[learningphysics]

Yes, that's right. Be careful about signs... show your calculations...

 

[UNG]

The answer is different since I did the Q4 and Q3.
It must be the signs.


F3x = (8.99 x10^9) x (200 x 10^-9) x (100 x 10^-9) / 0.05^2 x cos 90 i

+ (8.99 x10^9) x (200 x 10^-9) x ( -100 x 10^-9) / 0.0707^2 x cos 45 i

+ (8.99 x10^9) x (200 x 10^-9) x ( -200 x 10^-9) / 0.05^2 x cos 90 i

= 0 + -0.025 + 0 = -0.025





F3y = (8.99 x10^9) x (200 x 10^-9) x (100 x 10^-9) / 0.05^2 x sin 90 j

+ (8.99 x10^9) x (200 x 10^-9) x ( -100 x 10^-9) / 0.0707^2 x sin 45 j

+ (8.99 x10^9) x (200 x 10^-9) x ( -200 x 10^-9) / 0.05^2 x sin 90 j

= 71.92 + -25.44 x 10^-3 + - 143.84 x 10^-3 = 71.75

F3 = ( (-25 x10-3)^2 + (71.75)^2 ) Square Root
= (5148.06) Square Root
= 71.75

 

[learningphysics]

The angle between charge 3 and charge 4 is not 90... it is 0...

also, please be careful about signs... like I said before, get the magnitudes first, and then find out the sign separately by analyzing the situation... don't simply rely on the numbers to give you the right signs...

What is the direction of the charge number 2 exerts on number 3? don't use the numbers... just using the signs of the charges and the geometry you should be able to figure out the direction... is it upwards downwards to the left or to the right?

 

[UNG]

The direction of the charge that number 2 exerts on 3 will be downwards torwards the left because charge 3 is positive and charge 2 is negative. ?


The answers I get now from replacing 90 with 0 are,

F3x = 0 + - 0.025 + -143.84 x10^-3

so does this mean that -0.025 will become a +


Also for the direction of the charge that number 4 exerts on 3 will also be torwards the left because charge 3 is positive and charge 4 is negative. ?

This means that for, (when changing sin 90 to sin0)

F3y = 71.92 + -25.44 x 10^-3 + 0

-25.44 x 10^-3, Q2 on Q3 will stay negative.

Would that be it?

 

[learningphysics]

The direction of the charge that number 2 exerts on 3 will be downwards torwards the left because charge 3 is positive and charge 2 is negative. ?

charge 2 is attracting charge 3 since they are the opposite sign... if charge 3 is on the bottom left and charge 2 is on the top right... the force is upwards and to the right.

charge 1 repels charge 3 since they are the same sign... hence the force that 1 exerts on 3 is downwards...

charge 4 attracts charge 3 since they are the opposite sign... hence the force that 4 exerts on 3 is to the right.

The x-component of the force of 1 on 3 is:
0

The y-component of the force of 1 on 3 is:
-kq1q3/r^2 = 9*10^9(100*10^-9)(200*10^-9)/0.05^2 = -0.072N

So the force of 1 on 3 is: $$-0.072N\vec{j}$$

The x-component of the force of 2 on 3 is:
kq2q3/r^2*cos(45) = 9*10^9(100^-9)(200*10^-9)/0.0707^2cos(45) = 0.02546N

The y-component of the force of 2 on 3 is also: 0.02546N

So the force of 2 on 3 is: $$0.02546N\vec{i} + 0.02546N\vec{j}$$

The x-component of the force of 4 on 3 is: kq4q3/r^2 = 9*10^9(200*10^-9)(200*10^-9)/0.05^2 = 0.144N

The y-component of 4 on 3 is 0.

So the force of 4 on 3 is: $$0.144N\vec{i}$$

So the 3 forces are:

$$-0.072N\vec{j}$$
$$0.02546N\vec{i} + 0.02546N\vec{j}$$
$$0.144N\vec{i}$$

adding up the 3 forces gives the net force on 3:

$$0.16946N\vec{i} -0.04654\vec{j}$$

 

[UNG]

Thanks very much for your help!

I think I fully understand that question now
and have now managed to get the same answers.

Thanks

 

[learningphysics]

Thanks very much for your help!

I think I fully understand that question now
and have now managed to get the same answers.

Thanks

cool! no prob! glad you got the same answers, because I was making some mistakes, so I was a little worried that I might have messed up the answers. But I think everything's right now.