Calculate Net Force on Q3: Charge Position for Zero Net Force

[Winzer]

I assume you mean the distance from q3 to q2. r+x. Is that what you meant?

 

[rootX]

nopes,
but you are/ should be using q1 and q2?

 

[Winzer]

thought i am using q3 & q1

 

[rootX]

oh..that was wrong, I guess

 

[Winzer]

so what am I using?

 

[rootX]

q1, and q2

you are simply lost!, don't you? ^^

 

[Doc Al]

It seems that you are still a bit confused. Remember that you are trying to find out where to put Q3 so that it feels no net force from Q1 + Q2. Since Q3 feels the electric field from Q1 + Q2, this is equivalent to asking where is the net field from Q1 + Q2 equal to zero.

Let's look at the field to the left of Q1 at some point a distance X to the left of Q1. The field from Q1 at that point will be:
$$E_{q_1} = - \frac{k q_1}{x^2}$$

Note that I put a negative sign, since the field points to the left.

Now the field from Q2 (realize that if the distance from the point to Q1 is X, then its distance from Q2 must be X + 0.279 m):
$$E_{q_2} = + \frac{k q_2}{(x + 0.279)^2}$$

Note that this field points to the right, so it's positive.

Now add them up and solve for the distance (x) that makes the sum equal to zero.

Note that in both of the equations above, I let q_1 and q_2 stand for just the magnitude of the charges. (Don't put in a negative sign twice!)

 

[PFStudent]

The charges Q1= 1.90·10-6 C and Q2= -3.03·10-6 C are fixed at their positions, distance 0.279 m apart, and the charge Q3= 3.33·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

Hey,

In solving this problem refer to the following principle,

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Given any two arbitrary un-like sign charges: $q_{1}$ and $q_{2}$, placed on an axis a distance L from each other. Then, the placement (on that axis) of a charge $q_{3}$ such that the net force on $q_{3}$ due to: $q_{1}$ and $q_{2}$, will be zero. Can be given as follows,

$q_{1}q_{2} < 0$ $\therefore$ $q_{1}q_{2} \equiv -$

$$|q_{1}| < |q_{2}|, |\vec{r}_{31}| < |\vec{r}_{32}|, |\vec{r}_{32}| > L$$

$$|q_{1}| = |q_{2}|$$, No equilibrium exists on that axis.

$$|q_{1}| > |q_{2}|, |\vec{r}_{31}| > |\vec{r}_{32}|, |\vec{r}_{32}| > L$$
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I wrote up this principle up a while back, because these type of (electric charge) physics problems come up so often.

Thanks,

-PFStudent