- #1

sylent33

- 39

- 5

- Homework Statement
- Find the new position of Q3

- Relevant Equations
- Coloumbs Force

Hi!

Given three voltages as follows;

Q1 = 1C,Q2 = 1C,Q3 = 2C

The distance a is 1m and b = 2m

a) Find the values of the forces that are acting on Q2

I did that like this;

$$ F_{12} = \frac{Q1*Q2}{4\pi\epsilon r^2} $$

$$ F_{32} = \frac{Q1*Q3}{4\pi\epsilon r^2} $$

The results are : ##F_{12} = 8,99 GN and F_{13} = 4,49 GN ##

b) What is the direction and value of the total force as a sum of the individual forces. Here was also given a hint as follows:

"Think in terms of vectors"

My idea here was this: The force F12 is moving from Q1 to Q2,so its moving from left to right,the force F32 is moving from Q3 to Q2.Now they are going against each other,that would mean,in terms of vectors that I would have to subtract them from each other.Hence;

##F_{total} = 4,5 GN ## For the direction of Ftotal I am not sure.I think it should be from left to right since the force acting from the left (F12) is much greater (2x almost) than the force from the right (F32) so that would mean that it will "push out" and that Fg is poining to the right.

c) How would you need to chose a so that no force acts upon Q2

In order to be no force onto Q2 all the forces acting upon it have to be zero,hence;

$$ F_{12}+F{32} = 0 $$

$$ F_{12} = \frac{Q1*Q2}{4\pi\epsilon a^2} + F_{32} = \frac{Q2*Q3}{4\pi\epsilon r^2} = 0$$

since we want a,we need to move F32 to the other side.Than do some simplification;

$$ \frac{Q2}{a^2} = -\frac{Q3}{ r^2} $$

Plug in values

$$ \frac{1}{a^2} = -\frac{2}{ 2^2} $$

Solve for a

$$ a = \sqrt{-2} $$

And obviously this is not correct. The solution says it should be 1,41m which is just sqrt of 2.I am not sure how they get there,I've checked my algebra multiple times and it looks correct.It only leaves my initial idea to be at fault which is where I need a bit of help. How would you approach the problem c)?

Also what do you think about my solution/thought process in b) ?

Many thanks and greetings!

Given three voltages as follows;

Q1 = 1C,Q2 = 1C,Q3 = 2C

The distance a is 1m and b = 2m

a) Find the values of the forces that are acting on Q2

I did that like this;

$$ F_{12} = \frac{Q1*Q2}{4\pi\epsilon r^2} $$

$$ F_{32} = \frac{Q1*Q3}{4\pi\epsilon r^2} $$

The results are : ##F_{12} = 8,99 GN and F_{13} = 4,49 GN ##

b) What is the direction and value of the total force as a sum of the individual forces. Here was also given a hint as follows:

"Think in terms of vectors"

My idea here was this: The force F12 is moving from Q1 to Q2,so its moving from left to right,the force F32 is moving from Q3 to Q2.Now they are going against each other,that would mean,in terms of vectors that I would have to subtract them from each other.Hence;

##F_{total} = 4,5 GN ## For the direction of Ftotal I am not sure.I think it should be from left to right since the force acting from the left (F12) is much greater (2x almost) than the force from the right (F32) so that would mean that it will "push out" and that Fg is poining to the right.

c) How would you need to chose a so that no force acts upon Q2

In order to be no force onto Q2 all the forces acting upon it have to be zero,hence;

$$ F_{12}+F{32} = 0 $$

$$ F_{12} = \frac{Q1*Q2}{4\pi\epsilon a^2} + F_{32} = \frac{Q2*Q3}{4\pi\epsilon r^2} = 0$$

since we want a,we need to move F32 to the other side.Than do some simplification;

$$ \frac{Q2}{a^2} = -\frac{Q3}{ r^2} $$

Plug in values

$$ \frac{1}{a^2} = -\frac{2}{ 2^2} $$

Solve for a

$$ a = \sqrt{-2} $$

And obviously this is not correct. The solution says it should be 1,41m which is just sqrt of 2.I am not sure how they get there,I've checked my algebra multiple times and it looks correct.It only leaves my initial idea to be at fault which is where I need a bit of help. How would you approach the problem c)?

Also what do you think about my solution/thought process in b) ?

Many thanks and greetings!

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