14 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.Integrate field dx
Take x²+a² = t
Then dt = 2xdx
Multiply and divide by 2 to get 2xdx in the numerator
Woah woah woah what wait
Integrating field gives us potential?
Oh yea it does
Taking it t^2?
No, just t
Integrate t^(-3/2)
Cool
$\int_{\infty}^{x}\frac{Adt}{2t^{\frac{3}{2}}}$
Opt
Yo limits aisi kyu?
@Opt
Integrate, replace t for x²+a² then apply limits
Nono like why are taking from infinity to x
Because our reference for zero potential is at infinity
Oh makes sense
Thanks
@Opt check out the other one too https://discord.com/channels/1226379612238385242/1347826941494956113
+solved @Opt
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