E-Field in a Ring: Homework Solutions

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The E-field at the center of a uniformly charged metal ring is zero due to symmetry, as the electric fields from opposite sides cancel each other out. For points within the circumference of the ring, the E-field is not zero; it varies in magnitude and direction since the contributions from opposite points do not balance out. The confusion arises from the assumption that the E-fields at these points would cancel completely, which they do not. Understanding the vector nature of electric fields is crucial for analyzing points off-center. Clarification on these concepts can enhance comprehension of electric field behavior in symmetrical charge distributions.
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Homework Statement


a) What is the E-field at the center of a metal ring which has uniform charge density?

b) Is E=0 for any other point within the circumference of the ring?2. The attempt at a solution

For (a), since the ring has uniform charge density and is symmetrical, based on symmetry, E-field from opposite sides "cancels" each other out therefore the E field at the center of the ring should be E=0. Am I correct?

For (b), this is the one that got me confused. I was thinking that E-field at all other points (other than the center) should be non-zero as the E-fields from opposite ends at these points are opposite in direction however not equal in magnitude. However I am not too sure about this.

Greatly appreciate if anyone can enlighten me.

-yinx-
 
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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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