Distance between two charges where field strength = 0

In summary, the problem asked to find the point along the line connecting two charges where the total electric field due to the charges is equal to zero. Using the equation for electric field and setting it equal to zero, the variables were rearranged to find the distance from one of the charges to the point of zero electric field. After solving the equation, the correct distance was found to be 0.38 m.
  • #1
TFM
1,026
0
[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



[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]

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
 
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  • #2
From this point,
TFM said:
[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?
 
  • #3
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
 
  • #4
TFM said:
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.
 
  • #5
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
 
  • #6
TFM said:
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?
 
  • #7
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
 
  • #8
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!)
 
  • #9
TFM said:
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
 
  • #10
I just tried putting in 0.38, an it is the correct answer :smile:

Thanks everybody,

TFM
 

1. What is the formula for calculating the distance between two charges where the field strength is 0?

The formula for calculating the distance between two charges where the field strength is 0 is d = q1q2/4πε0, where d is the distance, q1 and q2 are the two charges, and ε0 is the permittivity of free space.

2. How is the distance between two charges related to the electric field strength?

The distance between two charges is inversely proportional to the electric field strength. This means that as the distance between the charges decreases, the electric field strength increases, and vice versa.

3. Can the distance between two charges ever be negative?

No, the distance between two charges cannot be negative. It is always a positive value because it represents the physical distance between the two charges.

4. How does the distance affect the strength of the electric field between two charges?

The distance between two charges has a significant impact on the strength of the electric field. As the distance increases, the strength of the electric field decreases. This is because the electric field spreads out as the distance increases, resulting in a weaker field.

5. What is the significance of the distance between two charges where the field strength is 0?

The distance between two charges where the field strength is 0, also known as the zero field point, is where the electric field is canceled out due to the equal and opposite charges. This point is significant because it helps us understand the behavior of electric charges and their interactions.

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