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.
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.
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
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
(edited 10 months 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
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 10 months 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
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.
Quick Reply
Related discussions
- Geography 12marker edexcel plzzz someone mark it and give advice
- Best books to read for Cambridge engineering?
- A level Physics AQA - Electricity question - not sure how to do
- Need some advice about future career. Please HELPPP
- OCR B A-level Physics Paper 3 Advancing Physics (H557/03) - 17th Jun 2023 [Exam Chat]
- Please help!
- OCR Physics GCSE Paper 1 (22nd May 2024)
- Is Physics A-Level even THAT hard?
- Computer Science or Further Maths for 3rd A-Level
- Integrated mechanical and electrical engineering vs pure mechanical
- Electrical Engineering VS Electrical and Electronic Engineering?
- Electrical Engineering with Math Bio Chem
- Electrical Eng at Bristol or Mechanical Eng at Sheffield
- are electrical and magnetic fields at right angles or opposite to eachother?
- AQA GCSE Physics Paper 1 (Foundation Tier triple) 8463/1F - 22nd May 2024 [Exam Chat]
- Engineering uni supercurriculars
- Civil engineering or Electrical/Electronic engineering Beng?
- Poor AS-Level Grades
- A level physics for engineering
- Physics paper 1 GCSE mock
Latest
Trending
Last reply 1 month ago
A level Edexcel physics aluminium foil passing through a uniform fieldPhysics
3
6
Last reply 1 month ago
If you could spin a hand spinner at the speed of light could it affect Space Time?Physics
6
5
Last reply 1 month ago
Using the equatorial diameter, how large is Mercury compared to Earth?Physics
2
3
