Calculating Electric Potential on an Infinite Charged Plane: A Homework Guide

AI Thread Summary
To calculate the electric potential difference between equipotential surfaces of an infinite charged plane with a uniform surface density of 1.00 nC, one must first determine the electric field generated by the plane. The electric field is constant and can be derived using Gauss's law, which is applicable despite the open nature of the plane. The potential difference of 10.0V can then be related to the separation between the equipotential surfaces using the relationship between electric field and potential. Understanding the scalar product of the electric field and displacement is crucial for this calculation. This approach provides a clear method to find the separation between the equipotential surfaces.
ibaraku
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Homework Statement



An infinite plane sheet of charge has uniform surface density
sigma = 1.00nC. What is the separation between equipotential surfaces differing in potential by 10.0V


I just a push in the right direction, thanks
 
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An infinite plane sheet of charge carries uniform electric field through infinite space excluding the sheet itself. Find the electric field; the electric potential should simply be the scalar product of the electric field and displacement (when finding the electric field, note that the Gaussian formula is for closed surfaces rather than open planes, or, by theory, surfaces that receive all electric flux/field lines...that is...if I have the concept right).
 
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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