How Do Surface Charge Densities Calculate on a Conducting Spherical Shell?

  • Thread starter Thread starter Trentonx
  • Start date Start date
  • Tags Tags
    Gauss's law Law
AI Thread Summary
The discussion focuses on calculating surface charge densities on a conducting spherical shell with a point charge at its center. A point charge of -4 µC induces a charge of +4 µC on the inner surface of the shell, while the remaining +1 µC of the shell's total charge of +5 µC resides on the outer surface. Participants express confusion about applying Gauss's law and the distribution of charges, clarifying that the presence of the inner charge affects the charge distribution. The conversation also touches on calculating the electric field at specific radii, indicating the relevance of Coulomb's Law and Gauss's Law in these calculations. Understanding these principles is essential for solving the problem accurately.
Trentonx
Messages
36
Reaction score
0

Homework Statement


A point charge of strength q1 = -4 µC is located at the center of a thick, conducting spherical shell of inner radius a = 2 cm and outer radius b = 3 cm. The conducting shell has a net charge of q2 = 5 µC
(a) Calculate the surface charge densities on the inner (sa) and outer (sb) surfaces of the spherical shell.


Homework Equations


Gauss's law
Surface area of sphere - 4*pi*r^2

The Attempt at a Solution


So I thought that on a conductor, the charge was all on the outside, so there would no charge on the inner surface, but that was wrong, so I'm at a lose on how to approach this. Gauss's law give me the electric field through an area, right? And then relates that to the charge enclosed, which I know? I might know the equation, but the concepts and how to apply them escape me.
 
Physics news on Phys.org
In the absence of the charge q1 at the center , all the charge q2 on the sphere would have been on the outer surface of the sphere. But due to the presence of q1, same amount of charge will be attracted towards inner surface of the sphere. The remaining charge will be on the outer surface of the sphere.
 
Alright, so I need to find the "spread" of charges, how much remained on the outside surface, and how much went towards the inside. How do I apply Gauss's law then? Can I assume that 4 µC of the sphere's charge went to the inner surface then to compensate for the charge inside?
 
So I found the charge densities, and it makes sense, as 4 µC were attracted to the inside of the sphere, since -4 µC were enclosed and left 1µC on the outer surface. Now, the follow up questions are asking about the electric field at various radii.
(b) Calculate the net radial electric field component at the following radii:
At r = 1cm?
At r = 6cm?

It seems this would be an application of Coulomb's Law or that E=(k)(Q)/(r^2), but these aren't the point charges that we have worked with. Gauss's Law tells me the net electric flux, right, so it doesn't seem to be useful in finding the electric field.
 
According to Gauss's law charged sphere acts like a point charge.
 
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...
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}...

Similar threads

Back
Top