Arranging Charged Objects to Exert Forces

[runningirl]

Homework Statement

1.) You have three charged objects: A; B; and C. Given that qA = 2.0 C; qB = - 3.0 C; and qC = -4.0 C, how could you arrange A, B, and C so that:
a. The force that A exerts on B is equal to .00675 N.
b. The net force exerted on C is 0.0 N.

Homework Equations

The Attempt at a Solution

..i have no idea where to start

 

[Bhumble]

$$\vec{F} = q \vec{E}, \vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{\hat{r}}{r^2}$$

You need to find a distance of A from B where they exert 0.00675 N on each other, which will be on a 1 dimensional axis, then add C somewhere along this line so that is equally attracted/repulsed and it cannot be between A and B since one of the objects repels it and the other attracts it.

 

[supratim1]

Hi Runningirl.

consider point A at origin. and B on x axis, with abscissa x. then solve the force equation. Now place C at X2 distance from A and (x + X2) from B, and solve to find X2. (keeping the above reason by Bhumble in mind).

 

[runningirl]

can i do .00675=9*10^9(2.0 )(3.0 )/d^2... find d for distance between a and b.
but how do i find the C distance?

 

[runningirl]

$$\vec{F} = q \vec{E}, \vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{\hat{r}}{r^2}$$

You need to find a distance of A from B where they exert 0.00675 N on each other, which will be on a 1 dimensional axis, then add C somewhere along this line so that is equally attracted/repulsed and it cannot be between A and B since one of the objects repels it and the other attracts it.

Sorry, but I don't understand where I could put C.

 

[supratim1]

consider C at X2. C cannot be between A and B. So once consider C on other side of A, next on the other side of B.

 

[runningirl]

does it matter how far away it is from a or b? because i said it didn't.

 

[supratim1]

of course it matters. it is affected by the electric fields of both A and B.

 

[runningirl]

right..
i understand that c has to be an X2 distance, but how do i find that x2 distance?

 

[Bhumble]

Two things to note is that superposition is applicable to the electric field and that the force on A from B is equal to the force on B from A.

For part (a) what you need to do is solve for r, which is the distance between the two point charges. $$\vec{F} = .00675 N = \frac{q_A q_B}{4 \pi \epsilon_0} \frac{\hat{r}}{r^2}$$

For part (b) it is essentially the same thing but you need to consider the distance of C from both A and B as well as determining the total force due to both of their individual forces. So
$$\vec{F} = 0 N = \frac{q_A q_C}{4 \pi \epsilon_0} \frac{\hat{r}}{r^2} + \frac{q_B q_C}{4 \pi \epsilon_0} \frac{\hat{r}}{r^2}$$
I didn't specify very well in the above formula but r is a different value in the two summations so when you solve the problem use better nomenclature than I did.

 

[SammyS]

So, have you found the distance that's needed between q_A and q_B?

Next, think about this:

The magnitude of the force that q_A exerts on q_C has to be equal to the magnitude of the force that q_B exerts on q_C. (The forces just need to be in opposite directions.)

So, should q_C be closer to q_A or to q_B?​

 

[runningirl]

shouldn't qc be closer to qa?
when i found the distance i said that d^2=some neg number so i think i did something wrong.
but i believe i had the right method.

.00675=9*10^9(2)(-3)/d^2