A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm

In summary, a 3.5 nC positive charge is located at the origin (0 cm), while a -10 nC negative charge is positioned at x = 2 cm.
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Homework Statement
A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm.

At what x-coordinate could you place a proton so that it would experience no net force?
Relevant Equations
(3.5/x^2)=(10/(.02+x)^2)
Apparently the answer is not 2.9 cm or 7.4mm. I've looked at similar questions in this thread and solving the exact same way gives me the same answers. I set it up as (3.5/x^2)=(10/(.02+x)^2).
What am I doing wrong?
 
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We cannot tell you what you did wrong if we don't know what you did. Please post your solution.
 
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  • #3
Count said:
Homework Statement: A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm.

At what x-coordinate could you place a proton so that it would experience no net force?
Relevant Equations: (3.5/x^2)=(10/(.02+x)^2)

Apparently the answer is not 2.9 cm or 7.4mm. I've looked at similar questions in this thread and solving the exact same way gives me the same answers. I set it up as (3.5/x^2)=(10/(.02+x)^2).
What am I doing wrong?
The distance squared to the -10 nC charge is not (0.02+x)^2 … even if x is in meters. (Please use appropriate units in equations)
 
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  • #4
Orodruin said:
The distance squared to the -10 nC charge is not (0.02+x)^2 … even if x is in meters. (Please use appropriate units in equations)
I don't understand. If 3.5nC is at the origin, and -10nC is 2cm to the right of that, then the only place for a point charge to be placed is to the left of the origin. That unknown distance is x, so the distance from -10nC to the proton is .02m+x. Solving for x would give the results I got, and my mistake may be in not using the negative sign to show the point charge to the left of the origin, so the answer would be -2.9cm. Is this not correct? If so, why not?
 
  • #5
Count said:
I don't understand. If 3.5nC is at the origin, and -10nC is 2cm to the right of that, then the only place for a point charge to be placed is to the left of the origin. That unknown distance is x, so the distance from -10nC to the proton is .02m+x. Solving for x would give the results I got, and my mistake may be in not using the negative sign to show the point charge to the left of the origin, so the answer would be -2.9cm. Is this not correct? If so, why not?
I agree with 2.9cm left of origin. You can rule out the other solution to the quadratic.
Taking x as distance left of the origin without saying so was a bit confusing.
 
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Count said:
and my mistake may be in not using the negative sign to show the point charge to the left of the origin,
That is your mistake. The statement of the problem asks "At what x-coordinate could you place a proton so that it would experience no net force?" Coordinates to the left of the origin are conventionally negative.
 
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FAQ: A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm

What is the electric field at a point midway between the two charges?

The electric field at the midpoint can be calculated by finding the contributions from both charges and then summing them vectorially. The distance from the midpoint to each charge is 1 cm (0.01 m). The electric field due to a point charge is given by \( E = \frac{k|q|}{r^2} \). For the 3.5 nC charge, the field at the midpoint is \( E_1 = \frac{(8.99 \times 10^9) \times 3.5 \times 10^{-9}}{(0.01)^2} \approx 3.15 \times 10^4 \) N/C directed away from the charge. For the -10 nC charge, the field at the midpoint is \( E_2 = \frac{(8.99 \times 10^9) \times 10 \times 10^{-9}}{(0.01)^2} \approx 8.99 \times 10^4 \) N/C directed toward the charge. Summing these fields (considering direction), the net electric field at the midpoint is \( E = 8.99 \times 10^4 - 3.15 \times 10^4 = 5.84 \times 10^4 \) N/C directed towards the -10 nC charge.

What is the potential difference between the two charges?

The electric potential due to a point charge is given by \( V = \frac{kq}{r} \). The potential at a point due to multiple charges is the algebraic sum of the potentials due to individual charges. At the origin (due to the 3.5 nC charge), \( V_1 = \frac{(8.99 \times 10^9) \times 3.5 \times 10^{-9}}{0} = \infty \), but this is not practical for a finite distance. For the -10 nC charge at 2 cm, \( V_2 = \frac{(8.99 \times 10^9) \times (-10) \times 10^{-9}}{0.02} \approx -4.495 \times 10^4 \) V. The potential difference \( V = V_2 - V_1 \approx -4.495 \times 10^4 \) V, considering practical distances.

What is the force between the two charges?

The force between two point charges is given by Coulomb's law: \( F = \frac{k|q_1 q_2|}{r^2} \). Here, \( q_1 = 3.5 \) n

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