Distance between two charges where field strength = 0

[TFM]

[SOLVED] Distance between two charges where field strength = 0

Homework Statement

Two particles having charges q1 = 0.500 nC and q2 = 9.00 nC are separated by a distance of 2.00 m.

At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

Homework Equations

$$E = \frac{1}{4 \pi \epsilon _0}\frac{q}{r^2}$$

I have reduced this two:

$$\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}$$

The Attempt at a Solution

putting in the variables:

$$\frac{0.5*10^{-9}}{r_1^2} = \frac{9.0*10^{-9}}{r_2^2}$$

I can rearrang to get:

$$\frac{r_2^2}{r_1_2} = \frac{9.0*10^{-9}}{0.5*10^{-9}}$$,

and:

$$\frac{r_2^2}{r_1_2} = 18$$

But I am not sure how to get the right distance to the actual point?

TFM

 

[Hootenanny]

From this point,

$$\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}$$

let x be the distance from the first charge to the point where E=0 such that,

$$\frac{q_1}{x^2} = \frac{q_2}{(2-x)^2}$$

Do you follow?

 

[TFM]

That makes sense, so now:

$$\frac{q_1}{x^2} = \frac{q_2}{4-4x+x^2}$$

Rearranging:

$$\frac{q_1}{q_2} = \frac{x^2}{4-4x+x^2}$$

And, inserting q1 and q2:

$$0.0556 = \frac{x^2}{4-4x+x^2}$$

How does this look?

TFM

 

[Hootenanny]

That makes sense, so now:

$$\frac{q_1}{x^2} = \frac{q_2}{4-4x+x^2}$$

Rearranging:

$$\frac{q_1}{q_2} = \frac{x^2}{4-4x+x^2}$$

And, inserting q1 and q2:

$$0.0556 = \frac{x^2}{4-4x+x^2}$$

How does this look?

TFM

Looks good to me, although I would be tempted to leave the ratio of the charges as a fraction.

 

[TFM]

So:

$$\frac{0.5*10^{-9}}{9*10^{-9}} = \frac{x^2}{4 - 4x + x^2}$$

Okay, so where should I go from here?

TFM

 

[Hootenanny]

So:

$$\frac{0.5*10^{-9}}{9*10^{-9}} = \frac{x^2}{4 - 4x + x^2}$$

Okay, so where should I go from here?

TFM

How about multiplying through by the denominator of the RHS, then solving for x?

 

[TFM]

So:

$$(0.5*10^{-9})(4 - 4x + x^2) = 9 x 10^{-9}x^2$$

$$(2*10^{-9} - (2*10^{-9})x + (0.5*10^{-9})x^2 = 9*10^{-9}x^2$$

Doubling everything:

$$(4*10^{-9} - (4*10^{-9})x + (1*10^{-9})x^2 = 18*10^{-9}x^2$$

$$(4*10^{-9} - (4*10^{-9})x + (1*10^{-9})x^2 = 18*10^{-9}x^2$$

Taking everything over to the LHS:

$$(4*10^{-9} - (4*10^{-9})x + ((1-18)*10^{-9})x^2 = 0$$

$$(4*10^{-9} - (4*10^{-9})x + (-17*10^{-9})x^2 = 0$$

Then using the formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where:

$$a = 4*10^{-9} , b = -4*10^{-9} , c = -17*10^{-9}$$

Giving:

$$x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{8*10^{-8}}$$

Which gives x as being either 2.62 (too big) or -1.62?

TFM

 

[Nick89]

You're correct except for the bit where you start using the quadratic formula.

The quadratic formula as you have written it corresponds to:
$$ax^2 + bx + c = 0$$

However, you're equation is in the form:
$$c + bx + ax^2 = 0$$

(Which is obviously the same, but you are switching a and c!)

 

[TFM]

Giving:

$$x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{8*10^{-8}}$$

Which gives x as being either 2.62 (too big) or -1.62?

TFM

Thanks, Nick89, This should be:

$$x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{2*-17*10^{-8}}$$

This gives -0.61 and 0.38, which look better

TFM

 

[TFM]

I just tried putting in 0.38, an it is the correct answer

Thanks everybody,

TFM