Potential between two particles

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The discussion centers on the interaction between two charged particles, +q and -q, highlighting that their potential is negative, indicating attraction. As the distance (r) between them approaches zero, the force of attraction increases, but the distance can never truly reach zero due to the physical size of the particles. When the charges come into contact, they may cancel each other out to some extent. Theoretical scenarios where particles are considered to have no size lead to contradictions, such as potential becoming infinite, which is not feasible in real physics. Ultimately, the conversation emphasizes that physical properties must be considered when discussing particle interactions.
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If we have two particles +q and -q. then the potential between them is <0 and therefore they will attract to each other. but what will happen when r is almost 0 (since we have r in the denominator)?
 
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The force of attraction gets larger as the distance diminishes.
The distance can never be zero because charges have size and the distance is measured from center to center. When the charges touch, some or all will cancel.
 


We cannot make the separation 0 between the two charges, however if you place a third particle very far away the force on the test particle from the difference in the charges will approximately cancel out.
 


I understand that, but let's say that the particles don't have "size", then in r=0 the potential will be infinity. the force of the attraction will cause them to get closer until r=0 , so how the potential can grow to infinty?
 


It can't because the particles do have size! You can't say "let's ignore physics" and then ask what physics says!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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