How Does Electric Field Vary with Distance in a Cylindrical Setup?

AI Thread Summary
The discussion focuses on calculating the electric field (E) in a cylindrical setup using Gauss's Law. For regions where the distance (r) is less than the inner radius (a), the electric field is determined to be E = λ / (2πεr). Between the inner radius (a) and outer radius (b), the electric field is zero due to the conducting nature of the shell. For distances greater than the outer radius (b), the electric field is calculated as E = (λ + λs) / (2πεr), accounting for the total charge enclosed. The calculations presented are confirmed to be correct.
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Homework Statement


based on the diagram attached, find the electric field E that varies with r.
the line and the metallic cylindrical shell are infinitely long.
the outer radius of the cylindrical shell is b and the inner one is a
and the line charge density for the line is \lambda
and the line charge density for the cylindrical shell is \lambdas

Homework Equations


Gauss's Law ( \ointE dA = Qin / \epsilon

The Attempt at a Solution


by using gauss's law with cylindrical gaussian surface coaxial with the line,
when r < a, i have E = \lambda / (2*pi*\epsilon*r);
for a< r < b, i have E = 0 since the shell is a conductor;

but for r > b, is it E = (\lambdas + \lambda) / (2*pi*\epsilon*r) since the total charge enclosed is the sum of charges in the two bodies?

Pls help me. thanks
 

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Your calculations are correct.
 
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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