Can someone answer a quick question for me about e field by spheres.

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For a uniformly charged solid sphere, the electric field inside the sphere (where R > r) is not zero and can be calculated using Gauss's Law. However, if the charge is uniformly distributed only on the surface of the sphere, the electric field inside is indeed zero. The confusion arises from mixing the concepts of solid spheres and spherical shells, as their electric field behaviors differ. In summary, the electric field depends on the charge distribution: it is zero inside a spherical shell but not inside a solid sphere. Understanding these distinctions is crucial for accurate calculations of electric fields.
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So let's say a sphere has radius R and you want to find the e field where R>r and R is equal to or greater than r.

would the e field for R>r just be 0?
Wouldn't that also mean it's 0 for R is equal to or greater than r, or does it change because it's EQUAL or greater?
 
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The answers depends on how the charge is distributed through the solid sphere.
 
sorry, uniformly distributed thorughout the sphere.
 
OK, in that case, the electric field wouldn't be 0 inside the sphere. You can calculate what it is by applying Gauss's Law.
 
i think i got the sphere confused with spherical shell because in the book it keeps switching back and forth. For spherical spheres it would be 0 right for points within the sphere?
 
Yes, if all the charge is on uniformly distributed on the surface of the sphere, there's no electric field inside.
 
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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