Potential difference inside a spherical shell

AI Thread Summary
The discussion focuses on calculating the potential difference between the north pole and the center of a uniformly charged spherical shell. The user attempts to apply Gauss's law to find the electric field (E) by integrating the surface charge density (sigma) over the shell's surface area. They express confusion about the left-hand side of their equation and whether to integrate E or the potential directly. The user concludes that E is uniform along the surface, leading to a simplified expression for E as sigma/epsilon0. The thread highlights the complexities involved in deriving the potential difference using both electric field and direct potential integration methods.
warfreak131
Messages
186
Reaction score
0

Homework Statement



A top half of a spherical shell has radius R and uniform charge density sigma. Find the potential difference V(b)-V(a) between point b at the north pole, and point a at the center of the sphere.

Homework Equations





The Attempt at a Solution



\oint E ds = \frac{\int \sigma ds}{\epsilon_{0}}

I integrated sigma ds from 0 to 2pi and 0 to pi/2, and got the total surface area as 2pi r^2. Now I want to solve for E, and then use -\int E dl to give me the potential difference, but what does the left hand side of my original equation equal?

E is uniform along the surface, so I can pull it out of the integral and be left with the integral of ds. But wouldn't that be 2pir^2 as well? I'd be left with E=sigma/epsilon0
 
Physics news on Phys.org
Would it be easier to integrate the potential directly?

V(r) = C * ∫ σ(r')*da'/[r - r']
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top