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Electric shapes physics

I have been struggling with this question https://isaacphysics.org/questions/electric_shapes?stage=all
I will show my working later, but essentially I equated the area of the triangle to the square so I got the length of the square. Then I equated k*l=k'*q_1^2/l^2 so I got q_1 in terms of other constants. Then I did the same for the square and got q_2 in terms of other constants. Nonetheless I am unsure wether this approach is incorrect. Maybe I should equate k*l? Maybe I should consider the 60° angle or maybe it is not k*l and k*l/2. Help would we greatly appreciated. :smile:
I just managed to solve it
basically what first you have to get an equation for length of triangle vs length of square so l vs x in my diagram.

equating forces for the triangle is lightwork -> the extension force = electric force
for the square you have to think about the diagonal electric force from the diagonally opposite charge and the electric force from the side opposite charge,
i.e. kx = F_d cos(theta) + F_s
then equating those forces will get you the answer.
Hope that helps :smile:
charge.png
(edited 1 month ago)
Thanks! However, why in the first part don't you do the same, considering the parallel force due to the top particle and then the other force?
In the triangle, say you consider the bottom right particle, the only forces acting on this particle are the extension force from the bottom left and top particle and the attractive force from the bottom left and top particle - since its in equilibrium these 2 forces must negate each other - in other words each has the same magnitude but opposite direction.

In the square, also consider the bottom right particle, the only forces acting on this particle are the extension force from the bottom left and top right particle and the attractive force from the bottom left and top right particle, BUT ALSO the diagonal force from the diagonally opposite particle.

If you split this diagonal force into its vertical and horizontal components, you get F_dsin(45) and F_dcos(45), since they are the same - this allows you to have symmetry along the diagonal so you only have to consider one "side" of the diagonal.

Hope that makes sense
(edited 1 month ago)
Original post by mosaurlodon
In the triangle, say you consider the bottom right particle, the only forces acting on this particle are the extension force from the bottom left and top particle and the attractive force from the bottom left and top particle - since its in equilibrium these 2 forces must negate each other - in other words each has the same magnitude but opposite direction.
In the square, also consider the bottom right particle, the only forces acting on this particle are the extension force from the bottom left and top right particle and the attractive force from the bottom left and top right particle, BUT ALSO the diagonal force from the diagonally opposite particle.
If you split this diagonal force into its vertical and horizontal components, you get F_dsin(45) and F_dcos(45), since they are the same - this allows you to have symmetry along the diagonal so you only have to consider one "side" of the diagonal.
Hope that makes sense

Thanks! It is true, in the triangle, the key is thinking about what foces are acting upon any particle you choose.

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