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I used the F = ((k)(q1)(q2))/r^2

F= ((9x10^9)(-14)(13))/(7.40^2))

F= 2.99 x 10^10

however that answer is wrong. (The correct answer is 177000000 N/C).

Which equation am I suppose to use?

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- Thread starter kbyws37
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In summary, the correct equation to use for finding the electric field at the midpoint between two point charges is E = ((k)(q))/r^2, where E is the electric field, k is the Coulomb constant, q is the charge, and r is the distance between the charge and the midpoint. When using this equation for two charges, the resultant electric field at the midpoint is the vector sum of the fields from each charge, E = E1 + E2. To ensure accuracy, it is important to be consistent with units and to properly convert between units if necessary.

- #1

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I used the F = ((k)(q1)(q2))/r^2

F= ((9x10^9)(-14)(13))/(7.40^2))

F= 2.99 x 10^10

however that answer is wrong. (The correct answer is 177000000 N/C).

Which equation am I suppose to use?

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- #2

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- #3

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hage567 said:The equation you have there is for the force between two point charges, not the electric field due to a point charge. Take another look at your notes and see if you can figure out which equation it should be. You need to consider the force due to each charge at the centre, i.e. the electric field at the centre due to Q1 and the force at the centre due to Q2. It is a vector quantity. Make sure you are consistent with your units.

So i would use the equation

F= ((k)(q))/r^2

so would i perform this equation twice? once for q1 and once for q2?

sorry, i am still not understanding

- #4

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Yes, that is the equation to use. You need to use it for both charges. The resultant electric field at the midpoint will be the vector sum of the fields from the two charges, E=E1+E2.

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hage567 said:

Yes, that is the equation to use. You need to use it for both charges. The resultant electric field at the midpoint will be the vector sum of the fields from the two charges, E=E1+E2.

I used the equation for both charges but still didn't get the right answer

E1 = ((9x10^9)(-14) / (7.4^2)) = - 2.3 x 10^9

E2 = ((9x10^9)(13) / (7.4^2)) = 2.14 x 10^9

E = (- 2.3 x 10^9) + (2.14 x 10^9) = 1.6

(correct answer should be 1.77 x 10^8)

- #6

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Also, you are not being consistent with the magnitudes of the quantities. 14uC needs to be 14x10^-6 C, so your units work out. Do you see what I mean? Include your units in the calculation so you can make sure they work out properly. Check your conversion from cm to meters as well (you must do this).

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thank you! you explained it very well and i won't forget converting my numbers! :)

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Electric field is a physical quantity that describes the force experienced by a charged particle in the presence of other charged particles. In the case of two point charges, the electric field is the force per unit charge experienced by a test charge in the vicinity of the two charges.

The electric field between two point charges can be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance between the two charges. Alternatively, the electric field can also be calculated by vector addition of the individual electric fields of the two charges.

The direction of the electric field between two point charges is always along the line connecting the two charges. It points away from a positive charge and towards a negative charge.

The magnitude of the charges has a direct impact on the electric field between two point charges. As the magnitude of the charges increases, the electric field increases as well. Conversely, a decrease in the magnitude of the charges results in a decrease in the electric field.

Yes, the electric field between two point charges can be zero. This occurs when the two charges are of equal magnitude and opposite sign, resulting in a net electric field of zero in between them.

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