15 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I understand that its in unstable equilibrium because if we display the particle P in any direction the net forces will no longer cancel out.
Can anyone send me the math behind this?
I know that i somehow have to show something using change in potential energy wrt distance/displacement (dU/dx) but idk where to even start
Source of question : Resnick Halliday chapter 37
First, calculate the potential as a function of x and y
How do i do that
Can you send me the solution pls? Im TERRIBLY lost atm
A hint would be fine as well
How would i go about expressing the potential as a function of x and y? I know that W = -dU/dx
Thats the only thing i can think of atm which might give me my answer
Ok, so, it's actually a lot simpler than you think
There's no need for the hard maths
Im thinking of W = -dU/dx since i can calculate the electro force
o
Consider a small displacement out of the plane
Along the z axis , assuming the charges are in xy plane
Alright
That's it.
Test charge will not return to P if pushed out of equilibrium perpendicular to the plans
Oh
Youre right
thank you bro
Idk how i didnt see that 💀
Sometimes there's a really simple solution and you become narrowsighted
yeah
the keyword in the question being
stable equilibrium
not just equilibrium
that's how you always check for stable or unstable equilibrium, see how it reacts to a small displacement/change+solved @Opt
Post locked and archived successfully!
Archived by
<@1035556259417571408> (1035556259417571408)
Time
<t:1742859548:R>
Solved by
<@763645886500175892> (763645886500175892)