Two charges on the x-axis

  • Thread starter derekbeau
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In summary: both wrong so idk if its because my equation for the electric field didnt work or if im doing something wrong with the calculations
  • #1
derekbeau
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I have two questions I am having trouble with, if you could please help me that would be great.

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Two point charges, q1 = 4.0×10^6 C and q2 = -1.0×10^6 C, are located on the x-axis at x1 = -1.0 cm and x2 = 3.0 cm

Determine the electric field at the origin

I have no idea how to do this
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Two protons are separated by a distance of 1.0×10^10 m (roughly the diameter of an atom). Calculate the electric force and gravitational force on one proton due to the other proton. The electric force is much larger. What is the ratio of the electric force to the gravitational force?

I thought id find the electric force by:
F = [k(1.6 x 10^-19)(1.6 x 10^-19)] / (10^-20) = 2.304 x 10^-8

And find the gravitational force by:
F = [(g m1 m2) / (10^-20)] = 2/742 x 10^-33

I don't know if those are right, but even if they are i don't know how to format the answer
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I just need some thorough explination on these two questions, i got all my other ones done.

Thanks
 
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  • #2
Q1
Because you were given that both point charges are along the x-axis essentially you are just being asked to find the net electric field resulting from the two point charges. Using your equation for the electric field you should be able to find the field for each point and from this the net field. Remember that the distance to the origin is not the same for both charges.

Q2
It's been a while but I believe you are right on. You calculated the electric and gravitational forces and indeed the electric force should be considerably stronger.
When setting up a ratio of the two forces it would be better to write it in the form Fe/Fg or Fg/Fe. Either way you will get the ratio of one force to the other.
 
  • #3
ok for Q1 i did

F = kQ / d^2 = [k(4 x 10^ -6)] / 1
F = kQ / d^2 = [k(-1 x 10^-6)] / 9

Then i added these two together but didnt get the correct answer
 
  • #4
nvm i got them
 

1. What is the electric field at a specific point on the x-axis due to two charges?

The electric field at a specific point on the x-axis due to two charges can be calculated using the formula E = k*q/(r^2), where k is the Coulomb's constant, q is the magnitude of the charge, and r is the distance from the point to the charges.

2. How does the distance between the two charges affect the electric field on the x-axis?

The electric field on the x-axis is inversely proportional to the square of the distance between the two charges. This means that as the distance increases, the electric field decreases, and vice versa.

3. What happens to the electric field at a point on the x-axis if the two charges have the same sign?

If the two charges have the same sign, the electric field at a point on the x-axis will be repulsive. This means that the electric field lines will point away from both charges.

4. Can the electric field at a point on the x-axis be zero?

Yes, the electric field at a point on the x-axis can be zero if the two charges have the same magnitude and opposite signs. This is known as an electric field equilibrium point.

5. How do the direction and strength of the electric field at a point on the x-axis change if the two charges have different magnitudes?

If the two charges have different magnitudes, the direction and strength of the electric field at a point on the x-axis will be determined by the larger charge. The electric field will point towards the larger charge and its strength will be greater than if the two charges had the same magnitude.

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