1Aragon could you be more clear on what you are trying to say ??
Three particles each of charge +q are placed on the vertices of a equilateral triangle of side 'a' .Now they all start moving away from each other symmetrically under the action of electrostatic force.What is the work done on each particle when they all move to infinity ?
Sol.A. Initial energy of the system was 3kq/a..final energy is 0...ΔE = 3kq2/a...(just forget about the signs for a moment)..Work done per particle is ΔE/3 = kq2/a..
Sol.B.. Initial energy associated with one particle is 2kq2/a...final energy is 0....so work done on each particle is 2kq2/a...
LOL :D
Which one of the above is correct and why ??
-
UP 0 DOWN 0 0 11
11 Answers
11Sol B is correct, isn't it? A charge can't attract or repel itself!
111No. I think both the solutions are wrong! We should first move one charge to ∞. Then move the second charge to ∞. No need to move the third charge. [1]
IGNORE REPLY IF IT SOUNDS STUPID[4]
3i think the soln A is correct. Actually the energy is conserved for the system which contains tthree charges.Because only mutual forces are acting.
1i think A coz... y shud we consider one charge wen they are all in one system...
we will consider the energy of the whole system .... initially n then finally...
1skygirl u dint explain why B is wrong if it has to be ? :D..what is the problem with considering one charge ? Give a proper reasoning.
9sol A is correct in sol B
ur neglecting that PE is shared btw the system
1hang on, y is kq/a evrwhere?? isn't it kq2/a ...
celestine's got a point but still how can u mathematically contradict B??
cn u plz post the solution??
1Thanks a lot to Bevin for pointing out my mistake of not writing q2 :D..Ok let me explain why B is wrong.
Work done = - ( change in potential energy) , if system is conservative ..
When we consider only once charge and the energy associated with it due to other charges we have to bear in mind that the other two charges are not stationary. So each charge moves in a non-stationary and non-potential field and hence work done won't be equal to change in potential energy.Sol A is correct when we take all 3 charges as system.
Always notice whenever u apply work done = - ( ΔU ) ...the field must be stationary or it will give erroneous results :D.