Please hlep me with this ELECTRICAL FORCE QUESTION.

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The discussion centers on calculating the net electrical force acting on a helium nucleus placed between two charged objects. The participant initially calculated the force as 1.12 X 10^-15 N but received the answer wrong, indicating confusion about the direction and magnitude of the forces involved. It was clarified that the alpha particle, being positively charged, experiences a repulsive force from the positive charge and an attractive force towards the negative charge. The correct approach involves recognizing that both forces act towards the negative charge, necessitating their addition for the net force calculation. Understanding the signs of the forces based on charge interactions is crucial for accurate results.
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5 An -=- particle (helium nucleus, 2 protons, and 2 nuetrons) is place between two stationary, charged objects as shown. It is 0.0030 meters to the right of the object carrying a +1.0 X 10^-12C charge and 0.0020 meters to the left of the object carrying a -2.0 X 10 ^-12 C charge.

a What is the net magnitude and direction of the electrical force acting on the particle?

I put 1.12 X 10^-15 N in the same direction and got the question WRONG.

b. At what rate and in what direction will the particle accelerate?

I put 3.4 X 10^11 m/s/s and also got it WRONG.

Any help would be greatly appreciated.

Im pretty sure I got the second one wrong since I got the 1st one wrong, and I have a feeling I got the 1st one wrong because I either added or subtraced wrongly, but I am confused about how I'm supposed to know the signs etc.
 
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Yes,the answer to the second part/question relies on te one to the first...
U'd have to add the 2 vectors which represent the electric forces which act on the ] \alpha particle...Can u do that??U need to chose an axis which goes through the 3 particles and a positive sense on it.Then u'd have to project the vector equation on this axis...

Daniel.
 
How did you get
1.12 X 10^-15 N
and what do you mean by
in the same direction

Which direction?
 
gnome said:
How did you getand what do you mean by

Which direction?
I found the force between the particle and the 2 objects and added them together. I found the force b/w the particle and the first object (leftmost) at 3.2^-16 and then found the force between it and the last object at -1.44X10^-15. Then I added them together. And as for the direction part, I probably just put that down because I had no real idea what it was talking about.
 
You forgot to consider that the direction of each force is related to the signs of the charges.


(Hint: what is the sign of the alpha particle's charge?)
 
Last edited:
gnome said:
You forgot to consider that the direction of each force is related to the signs of the charges.


(Hint: what is the sign of the alpha particle's charge?)
The alpha particle is postive so the force between it and the first object would be positvie too right? and the force between it and the second object would be negative? And I would have to add these together?
 
Which force is "positive" and which is "negative" depends on how you define your coordinates.

So the way I would state it is:
The charge of the alpha particle is positive so there is a repulsive force between it and the stationary positive charge, and an attractive force between it and the stationary negative charge. So both forces act in the direction towards the negative charge, and therefore, yes, you have to add them together.
 
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