Expression for Electric Field Outside Sphere

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The discussion centers on understanding the electric field outside a sphere using Gauss's law. Participants express confusion over a question that describes a sphere as both uniformly charged and having a variable charge density, which is contradictory. Clarification is sought on the intended meaning of the question. The importance of making an effort to solve the problem before seeking help is emphasized. Overall, the conversation highlights the need for precise language in physics problems.
eric11201120
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Homework Statement
The charge desnsity within a sphere of radius R is ρ = ρ0 - ar2, where ρ0 and a are constants and r is the distance from the center. Find an expression for a such that the electric field outside the sphere is zero.
Relevant Equations
Electric Field outside a sphere
E = KQ/r2
Electric flux
ΦE=∫EdA
I'm not quite sure where to start. If someone could help me I would very much appreciate it.
 
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Hint 1: What does Gauss's law tell you?
Hint 2: Given that expression for charge density, what's the total charge within the sphere?
 
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Steve4Physics said:
But there seems to be a mistake in the question. The sphere can't be both 'uniformly charged' and have a charge density which is a function of r.
Yes, a very sloppily worded question!
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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