How Does Changing Distance Affect Repulsive Force Between Charged Particles?

[Kathi201]

Charges Q1 and Q2 exert repulsive forces of 10N on each other. What is the repulsive force when their separation is decreased so that their final separation is 80% of their initial separation.
A. 16N
B. 12N
C. 10N
D. 8.0N
E. 6.4N

I think you have to use the equation F = K (Q1)(Q2)/ r^2 and multiply the distance (r) by .80 but I'm not sure how you calculate the force if you also have another unknown in the equation (r).

Another question..

An electron is moving horizontally east in an electric field that points vertically upward. The electric force on the electron is
A.zero
B.upward
c.downward
d.to the west
e. to the east

My thoughts are that it is B. upward. The reason for my thinking is because I am pretty sure the electric field points in the same direction as the force but I am not quite sure if my thinking is right.

any help would be really appreciated...

 

[edziura]

For the 1st question, if ri is the initial separation, then rf (final separation) is 0.8ri. The ri will cancel out.

Your answer to the second question is correct.

 

[Deadleg]

bleh edziura would be right.

 

[Kathi201]

Thank you for your help. I am still a bit confused on the first problem. If I use the equation F = K (Q1)(Q2)/r^2 I am not getting any of the answers that are provided. Here's how I worked it out

F = 8.99E9 (10N)(10N) / (Rf - Ri)^2

Rf = .8Ri

F = 8.99E9 (10N)(10N) / (.8Ri - Ri)^2 so Ri cancels

I have a feeling that my Q1 and Q2 values are wrong but I am not sure where to go from here.

Thanks again for all of your help!

 

[Deadleg]

$$Q$$ is not force, it is charge. They are simply constants in the equation in this context. Try using $$F\propto\frac{1}{r^2}$$

Shouldn't the second question be downwards, because if the field lines goes upwards, that means a negative charge will be at the top, therefore repelling the electron downwards?

 

[Kathi201]

So for the first question if F is proportional to 1/r^2 than this is what I am getting

Ff-10N = 1/r^2

Ff-10N = 1/.8^2

Ff-10N = 1.56

Ff = 1.56 +10N = 11.6 or 12N which is one of my answers.


For the second question I came up with the answer upward by looking at the equation F = qE. This shows that whatever direction the electric field is in, which in this case is vertically upward the electric force on that electron is upward. Your conclusion could be right but I think this is what we talked about in class this week.

Thanks again for your help!

 

[Kathi201]

Now that I think about it I think you might be right for the second question because an electron is negatively charged and if there is a negatively charged object the force and the electric field are opposite. Thanks for the tip on that one!

 

[Deadleg]

And for an electron, $$q$$ is negative, so in $$F=Eq$$, $$F$$ will be negative ie against the electric field :)

I don't think proportionalities work like that... I haven't done a lot of work with them, but by looking at the relationship, if r decreases, $$F$$ will increase yes? So if r decreases by 80%, $$r^2$$ decreases by 64%, then $$F$$ must increase by the same percentage.