Hard Electric Force Problems

[I Like Pi]

Homework Statement

There are two questions, one being a communicational question while the other an application:

1.Charge A of +2.0 muC is put at the origin of an cartisian plane. Charge B of -4.0 muC is put 40 cm to the right of charge A. Where should charge C be placed (between the two, to the left of A or to the right of B) to have a net force of zero? Argue without working out a solution. Do this for both a positive charge C and negative charge C.

2.Charge A of +5.7 muC and charge B of -3.5 muC are 25 cm apart. Where would you put charge C so it has no net force?

*keep number 2 to a linear one

Homework Equations

The Attempt at a Solution

For 1, I think that the third charge would go to the left of A no matter the sign (positive or negative) because if it was positive the +2 would repel it, but the -4 (since it is larger) would attract it. If the third charge were to be negative, the +2 would attract, but -4 (since it is large) would repel it back. Paraphrase: in both cases there would be a net force of 0, theoretically.

But i am not sure, because what if the magnitude of the thrid charge is much larger than 4... does it change

For 2, I really have no clue.. wouldn't you have to know the third charge?.. Would the third charge have to be between the two?..

Thanks for your time

 

[tiny-tim]

Hi I Like Pi!

For 1, I think that the third charge would go to the left of A no matter the sign (positive or negative) because if it was positive the +2 would repel it, but the -4 (since it is larger) would attract it. If the third charge were to be negative, the +2 would attract, but -4 (since it is large) would repel it back. Paraphrase: in both cases there would be a net force of 0, theoretically.

But you haven't yet explained why the charge can't be to the right.

But i am not sure, because what if the magnitude of the thrid charge is much larger than 4... does it change

No, if the force is to be zero, then the field must be zero, so it doesn't matter what the charge is, does it?

(Same for problem 2.)

 

[I Like Pi]

Hi I Like Pi!


But you haven't yet explained why the charge can't be to the right.

Well it can't be between A and B because charge A and B don't have the same magnitude.. And it can't be to the right of B because B is much more stronger than A, so if C is positive and to the right of B, B would cause it to attract and A isn't strong enough to repel it back to create a net force of 0.. vice versa..

Or am I wrong?

No, if the force is to be zero, then the field must be zero, so it doesn't matter what the charge is, does it?

Do you mean electric field? As in E = kq/r^2?

 

[tiny-tim]

Well it can't be between A and B because charge A and B don't have the same magnitude.. And it can't be to the right of B because B is much more stronger than A, so if C is positive and to the right of B, B would cause it to attract and A isn't strong enough to repel it back to create a net force of 0.. vice versa..

Or am I wrong?

No, you are right, but in an exam question you have to say all that!

Do you mean electric field? As in E = kq/r^2?

Yup!

 

[I Like Pi]

Thank you and you're right! haha..

Though for 2, would I have kq_A/r²=kq_B/(25-r)²?
Cause I can't use F=kq_Aq_B/r²?

Thanks Tim, you rock!

 

[tiny-tim]

Though for 2, would I have kqA/r^2=kqB/(25-r)^2?
Cause I can't use F=kqAqB/r^2?

(try using the X² and X₂ icons just above the Reply box )

Yes, though of course they're different "A"s … the "B" is the charge of the test (third) particle, which you don't bother to include because it cancels.

 

[I Like Pi]

(try using the X² and X₂ icons just above the Reply box )

Yes, though of course they're different "A"s … the "B" is the charge of the test (third) particle, which you don't bother to include because it cancels.

I don't get what you mean... I'd use the field equation and not the force?

Sorry for the confusion...

 

[tiny-tim]

If you leave out one of the charges (or if you use the value 1 for it), then you are using the field instead of the force …

if you want to keep the force, you need to put in a value for the "test" charge (and cancel it soon after you put it in).

 

[I Like Pi]

If you leave out one of the charges (or if you use the value 1 for it), then you are using the field instead of the force …

if you want to keep the force, you need to put in a value for the "test" charge (and cancel it soon after you put it in).

Sounds good, if I use force equation, i get this:

F_AC = F_CB

kq_Aq_C/r²=kq_Bq_C/(.25-r)²

q_Aq_C/r²=q_Bq_C/(.25-r)²
q_Aq_C(.25-r)²=q_Bq_Cr²
q_A(.25-r)²=q_Br²
q_A/q_B = r²/(.25-r)²

So then sub in the values..

5.7*10^-6/-3.5*10^-6 = r²/(.0625-.50r+r²)
-1.6285714(.0625-.50r+r² = r²
-.101785714 + .814285714r -1.6285714r² = r²
-.101785714 + .814285714r -2.6285714r² = 0

then use quadratic formula, but it has no real roots...

 

[tiny-tim]

5.7*10^-6/-3.5*10^-6 = r²/(.0625-.50r+r²)

No, it's 3.5 not minus 3.5 …

your formula already assumed that the two charges had opposite values …

you should get a result with the correct sign ​

 

[I Like Pi]

No, it's 3.5 not minus 3.5 …

your formula already assumed that the two charges had opposite values …

you should get a result with the correct sign ​

I just figured that out! Thanks tim! I totally forgot that the negative sign meant repulsive! Haha, I end up getting -1.41 and .11. Therefore the third charge would be placed about 11 cm to the right of charge A?

And I was just wondering, do you have to put the third charge between the two? And is that because you don't know the magnitude of the third charge?

Thanks again, tim!

 

[tiny-tim]

And I was just wondering, do you have to put the third charge between the two? And is that because you don't know the magnitude of the third charge?

Sorry, you've completely lost me …

I thought you just proved that it had to be on the right?

(and, as I said before, it doesn't matter what the third charge is)

 

[I Like Pi]

Sorry, you've completely lost me …

I thought you just proved that it had to be on the right?

(and, as I said before, it doesn't matter what the third charge is)

well yes, I proved it to be to the RIGHT of A, but it's because i modeled it that way, no? I used .25-r, meaning it had to be between A and B unlike my first question where the third charge was to the left of A...

 

[tiny-tim]

oh i see …

no, when i saw your (.25-r)², I assumed you meant (r-.25)², which is the same thing …

so it turned out ok! ​

 

[I Like Pi]

oh i see …

no, when i saw your (.25-r)², I assumed you meant (r-.25)², which is the same thing …

so it turned out ok! ​

hey, my teacher said i was doing it wrong :( how could it be wrong... apparently it can't be inbetween the two charges...

 

[tiny-tim]

hey, my teacher said i was doing it wrong :( how could it be wrong... apparently it can't be inbetween the two charges...

do you mean your teacher said it can be or it can't be between the two charges?

because it can't be

anyway … let's have another look …

2.Charge A of +5.7 muC and charge B of -3.5 muC are 25 cm apart. Where would you put charge C so it has no net force?

*keep number 2 to a linear one

q_Aq_C/r²=q_Bq_C/(.25-r)²
q_Aq_C(.25-r)²=q_Bq_Cr²
q_A(.25-r)²=q_Br²
q_A/q_B = r²/(.25-r)²

So then sub in the values..

5.7*10^-6/-3.5*10^-6 = r²/(.0625-.50r+r²)
-1.6285714(.0625-.50r+r² = r²
-.101785714 + .814285714r -1.6285714r² = r²
-.101785714 + .814285714r -2.6285714r² = 0

then use quadratic formula, but it has no real roots...

well, the method looks ok to me, except that it's a bit long-winded, and it would have saved time to stop at q_A(.25-r)²=q_Br²

 

[I Like Pi]

do you mean your teacher said it can be or it can't be between the two charges?

because it can't be

anyway … let's have another look …




well, the method looks ok to me, except that it's a bit long-winded, and it would have saved time to stop at q_A(.25-r)²=q_Br²

he got an answer of .95 m to the right of the negative charge... :/

said it's NOT inbetween... how

 

[tiny-tim]

he got an answer of .95 m to the right of the negative charge... :/

said it's NOT inbetween... how

yes, it can't be between, for the same reason as problem 1 …

i thought you understood that?

let's see your equation to see why you got .11 cm instead of .95 cm (ie r = .36 instead of 1.20)

 

[I Like Pi]

yes, it can't be between, for the same reason as problem 1 …

i thought you understood that?

let's see your equation to see why you got .11 cm instead of .95 cm (ie r = .36 instead of 1.20)

please... i am so confused :(

like i know it can't be in the middle...