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Distance between two charges where field strength = 0

  • Thread starter TFM
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TFM
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[SOLVED] Distance between two charges where field strength = 0

1. 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?

2. Homework Equations

[tex] E = \frac{1}{4 \pi \epsilon _0}\frac{q}{r^2} [/tex]

I have reduced this two:

[tex]\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}[/tex]

3. The Attempt at a Solution

putting in the variables:

[tex]\frac{0.5*10^{-9}}{r_1^2} = \frac{9.0*10^{-9}}{r_2^2}[/tex]

I can rearrang to get:

[tex]\frac{r_2^2}{r_1_2} = \frac{9.0*10^{-9}}{0.5*10^{-9}}[/tex],

and:

[tex]\frac{r_2^2}{r_1_2} = 18 [/tex]

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

TFM
 

Answers and Replies

Hootenanny
Staff Emeritus
Science Advisor
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From this point,
[tex]\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}[/tex]
let x be the distance from the first charge to the point where E=0 such that,

[tex]\frac{q_1}{x^2} = \frac{q_2}{(2-x)^2}[/tex]

Do you follow?
 
TFM
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That makes sense, so now:

[tex]\frac{q_1}{x^2} = \frac{q_2}{4-4x+x^2}[/tex]

Rearranging:

[tex]\frac{q_1}{q_2} = \frac{x^2}{4-4x+x^2}[/tex]

And, inserting q1 and q2:

[tex] 0.0556 = \frac{x^2}{4-4x+x^2}[/tex]

How does this look?

TFM
 
Hootenanny
Staff Emeritus
Science Advisor
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That makes sense, so now:

[tex]\frac{q_1}{x^2} = \frac{q_2}{4-4x+x^2}[/tex]

Rearranging:

[tex]\frac{q_1}{q_2} = \frac{x^2}{4-4x+x^2}[/tex]

And, inserting q1 and q2:

[tex] 0.0556 = \frac{x^2}{4-4x+x^2}[/tex]

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
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So:

[tex]\frac{0.5*10^{-9}}{9*10^{-9}} = \frac{x^2}{4 - 4x + x^2}[/tex]

Okay, so where should I go from here?

TFM
 
Hootenanny
Staff Emeritus
Science Advisor
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So:

[tex]\frac{0.5*10^{-9}}{9*10^{-9}} = \frac{x^2}{4 - 4x + x^2}[/tex]

Okay, so where should I go from here?

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

[tex] (0.5*10^{-9})(4 - 4x + x^2) = 9 x 10^{-9}x^2 [/tex]

[tex] (2*10^{-9} - (2*10^{-9})x + (0.5*10^{-9})x^2 = 9*10^{-9}x^2 [/tex]

Doubling everything:

[tex] (4*10^{-9} - (4*10^{-9})x + (1*10^{-9})x^2 = 18*10^{-9}x^2 [/tex]

[tex] (4*10^{-9} - (4*10^{-9})x + (1*10^{-9})x^2 = 18*10^{-9}x^2 [/tex]

Taking everything over to the LHS:

[tex] (4*10^{-9} - (4*10^{-9})x + ((1-18)*10^{-9})x^2 = 0 [/tex]

[tex] (4*10^{-9} - (4*10^{-9})x + (-17*10^{-9})x^2 = 0 [/tex]

Then using the formula:

[tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

where:

[tex] a = 4*10^{-9} , b = -4*10^{-9} , c = -17*10^{-9} [/tex]

Giving:

[tex]x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{8*10^{-8}}[/tex]

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

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

The quadratic formula as you have written it corresponds to:
[tex]ax^2 + bx + c = 0[/tex]

However, you're equation is in the form:
[tex]c + bx + ax^2 = 0[/tex]

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

[tex]x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{8*10^{-8}}[/tex]

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

TFM
Thanks, Nick89, This should be:

[tex]x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{2*-17*10^{-8}}[/tex]

This gives -0.61 and 0.38, which look better

TFM
 
TFM
1,026
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I just tried putting in 0.38, an it is the correct answer :smile:

Thanks everybody,

TFM
 

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