Force on Electron Due to Two Charged Objects in an Electric Field

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The discussion centers on calculating the electric field at Point A, located near two charged objects, Q1 and Q2. The electric field magnitude at Point A is found to be 5.265E3 N/C directed to the right. To determine the force on an electron placed at Point A, the formula F = qE is applied, resulting in a force of 8.42E-16 N directed to the left. Clarifications are provided on using Coulomb's law and the relationship between electric field and force. The conversation concludes with a confirmation of the correct approach to solving the problem.
blue_lilly
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Homework Statement


Two charged objects are separated by a distance of L=0.4650 m as shown in the diagram.
Object Q1 has a charge of +5.250 nC. Object Q2 has a charge of +2.960 nC.

A) What is the magnitude and direction of the electric field at Point A, which is located a distance d=0.09300 m to the right of Object Q1?

B) If you were to place an electron at Point A, what would be the magnitude and direction of the force on the electron?

Homework Equations


E= k (q/r^2)

F= (k*Q1*Q2)/ (r^2)

The Attempt at a Solution



A) [ (8.99E9) ((5.25E-9)/(.09300^2) ] - [ (8.99E9) ((2.96E-9)/(.372^2) ]=
= 5265 N/C = 5.265E3 N/C to the right THIS ANSWER IS CORRECT

B) The charge on an electron is -1.6E-19. I am also assuming that i will need Coulombs Law to solve.
[(8.99E9)(5.25E-9)(2.96E-9)] / (.4650) =.64610E-7 to the left
This is not the right answer and I'm not sure why. AM i even using the right formula?

Any help would be greatly appreciated. Thanks!
 

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You could use Coulomb's law and work directly with the charge values (you'd need to consider all the charges involved). But since you've already determined the electric field at point A you can use a different formula which relates the force on a charge due to the electric field...
 
gneill said:
You could use Coulomb's law and work directly with the charge values (you'd need to consider all the charges involved). But since you've already determined the electric field at point A you can use a different formula which relates the force on a charge due to the electric field...

The only equation I have that relates force and charge is E= F/q
(5.265E3)= F/(-1.6E-19)
(5.265E3)(-1.6E-19)= F
F= 8.42E-16 N to the left

Was that the equation you were talking about?
 
blue_lilly said:
The only equation I have that relates force and charge is E= F/q
(5.265E3)= F/(-1.6E-19)
(5.265E3)(-1.6E-19)= F
F= 8.42E-16 N to the left

Was that the equation you were talking about?

Yup. Most often you'll see it in the arrangement F = qE. It's analogous to the formula for force due to a gravitational field: F = mg. Makes it easy to remember :smile:
 
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gneill said:
Yup. Most often you'll see it in the arrangement F = qE. It's analogous to the formula for force due to a gravitational field: F = mg. Makes it easy to remember :smile:

Oh, OK! That makes sense. Thank you for your help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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