Net force on particle 1 due to particle 2 in vector form?

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The discussion focuses on calculating the net force on particle 0 due to particle 1, both of which are positively charged. The user expresses difficulty in obtaining the correct answer for their quiz on MasteringPhysics. The relevant equation, Coulomb's Law, is provided, indicating that the force can be expressed as a vector involving constants k, charges q0 and q1, and the distance d1. The direction of the force is emphasized, with the force on particle 0 acting in the direction away from particle 1. Understanding the unit vector from particle 1 to particle 0 is crucial for determining the correct vector form of the force.
Corey Bacon
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Homework Statement


Hi all,
I have this quiz on MasteringPhysics, but I can't seem to get the right answer.[/B]
Consider two positively charged particles, one of charge q0 (particle 0) fixed at the origin, and another of charge q1 (particle 1) fixed on the y-axis at (0,d1,0). What is the net force F⃗ on particle 0 due to particle 1?
Express your answer (a vector) using any or all of k, q0, q1, d1, i^, j^, and k^.

If someone wouldn't mind giving me some advice, I would much appreciate it.
Thanks

Homework Equations


Columns Law

The Attempt at a Solution


$$\frac{kq_0q_1}{d_1^2}j$$
 
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The direction of the force is important too.
 
Force on q0 by q1 acts in which direction ?
 
You have write down the part of the answer.
The complete Coulomb law is:
$$\vec{F_{12}}=\frac{kq_1q_2}{r_{12}^2}\vec{e_{12}}$$
for ##\vec{F_{12}}## is the force ##q_1## makes on ##q_2## and ##e_{12}## is the unit vector from ##q_1## toward ##q_2.##
It may help you find out the correct direction of the force.
 
41643ec5136542b93edf3f6b56fb90f0.png
and putting r(cap) = r(vector)/r
upload_2015-8-4_16-27-0.png
 
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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