Does Zero Electric Potential Necessarily Imply Zero Electric Field?

Thread starter lockedup
Start date Sep 12, 2009
Tags

Electric Electric field Field Potential

AI Thread Summary

Zero electric potential (V = 0) at a point does not necessarily imply that the electric field (E) is also zero, as potential energy may have converted to kinetic energy, indicating a non-zero electric field. Conversely, if the electric field is zero at a point, it does imply that the electric potential is constant, but not necessarily zero. The relationship between electric field and electric potential is defined by the gradient, where E is the negative gradient of V. Understanding this relationship is crucial for solving problems involving electric fields and potentials. The discussion emphasizes the need to clarify the implications of zero values in electric potential and field contexts.

Sep 12, 2009

lockedup

Homework Statement

If V = 0 at a point in space, must E = 0? If E = 0 at some point, must V = 0 at that point?

Homework Equations

The Attempt at a Solution

V = 0 means there is no PE at that point is zero. However, the PE may have been converted to KE which would require a non-zero E field.

Am I in the ballpark? Now, I just need the second part...

Physics news on Phys.org

Sep 12, 2009

rl.bhat

Homework Helper

What is the relation between E and V?

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

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}...

View full post »

Thread 'Understanding Forces and their Vector Components'

I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?

View full post »