Potential Energy of a Proton Near a Positively Charged Sphere: Does it Decrease?

AI Thread Summary
As a proton approaches a positively charged sphere, there is a debate about its potential energy. One viewpoint suggests that the potential energy decreases due to the repulsive forces between the proton and the sphere. Conversely, another argument states that the potential energy increases because work is required to move the proton closer to the sphere. The discussion highlights the complexity of potential energy in electrostatic contexts. Ultimately, understanding the interplay of forces is crucial in determining the behavior of the proton's potential energy.
daniel69
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A proton is brought near a positively charged sphere. As it is brought closer its potential energy...

A. increases
B. decreases
C. remains the same
D. cannot determine

Does it decrease (B) because same forces repel causing a decrease in PE?
 
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I would say that it increases because you need to do some work in order to bring the proton near the positively charged sphere.
 
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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