Charges placed on a hexogon, working out the force and energy

AI Thread Summary
The discussion focuses on calculating the net force acting on a charge at point D in a hexagon with six equal charges and the total electrostatic energy of the system. Participants emphasize using symmetry to simplify calculations, suggesting that only the forces from the bottom half of the hexagon need to be considered. For the energy calculation, it is recommended to sum the contributions from each charge as they are added to the system, while being cautious to avoid double-counting. The importance of symmetry in physics problems is highlighted, particularly in electrostatics. Overall, symmetry is presented as a powerful tool for simplifying complex calculations in physics.
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Six equal charges, each Q coulombs, are placed at the vertices of a regular
hexagon of side a, as shown above. What is the net force acting on the charge
at D as a result of the other 5 charges? [3]
d) Calculate the total electrostatic energy of the system of 6 charges in (c)

Its only work 3 marks so i thought resolving all the forces for each charge would be a be a bit long and was wondering if there was a quicker method ?

also how would you approach the second part of the question (d)

Cheers

Heres a photo of the question

http://i1200.photobucket.com/albums/bb327/loba333/photouckt.png
 
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Look at the symmetry of the problem is there anything that you can eliminate from the symmetry?
 
Im new to the whole symmetry concept in terms of using to help with physics problems.

are u saying i only need to work out the bottom half of the hexagon ie the force due to A F and E, then i can just use the same forces and say E=C and F=B but with opposite y coponets ie j=-j
 
are u saying i only need to work out the bottom half of the hexagon ie the force due to A F and E

Yes! That is exactly what I'm saying. However furthermore:

then i can just use the same forces and say E=C and F=B but with opposite y coponets ie j=-j

If the y-components of E and F are the same as C and B except with "-j", then what will you get if you add the y-components of just these four charges (E_y+C_y+F_y+B_y = ?) together?
 
also for the second part how would u calculate the electrostatic energy of the entire system ?

i know you would use U=kq1q2/r

But which ones do i use. I know from a lecture it is easy to double the energy unnecessarily .
 
loba333 said:
also for the second part how would u calculate the electrostatic energy of the entire system ?

i know you would use U=kq1q2/r

But which ones do i use. I know from a lecture it is easy to double the energy unnecessarily .

You can sum the contributions to the overall energy that occur when you bring each charge in one by one. As each one "arrives", you calculate the work done against each of the other charges already in place (so the first one's "free"). For six charges that's a fair amount of algebra. You may want to pre-calculate all the distances from one vertex to all the others (again, symmetry helps) for reference.

Keep in mind that sin(30) = 1/2, and cos(30) = sqrt(3)/2; You don't have to carry the sin and cos functions along through the calculations.
 
loba333 said:
Cheers man, symetrys a pretty beautiful thing when your doing somthing like this.
I'v just worked through it, would u mind having a look to see if its right

Kind regards

Dave

http://i1200.photobucket.com/albums/bb327/loba333/IMG_1459.jpg

The answer looks good! Symmetry (not just necessarily reflectional symmetry as this problem demonstrates) is probably the most amazing idea you will ever learn about in physics.

But the take home message for this problem is that in the case of electro/magneto-statics typically you always want to start by looking for symmetry in problems.
 
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