Request Assistance with Gauss's Law- Surface charge of sphere

AI Thread Summary
A 240nC point charge at the center of an uncharged spherical conducting shell induces a surface charge on the inner surface to maintain zero electric field within the shell, resulting in an equal and opposite charge on the outer surface. The surface charge density can be calculated using the formula for charge density, which is the total charge divided by the surface area of the shell. The electric field strength at the outer surface can be determined using the formula E = q/(4*pi*E*r^2). The initial calculations for surface charge density and electric field strength provided by the user were incorrect, but the guidance indicates that understanding the relationship between the charges is crucial. This discussion emphasizes the importance of grasping fundamental concepts in electrostatics for solving problems involving Gauss's Law.
chipperh
Messages
5
Reaction score
0

Homework Statement


a 240nC point charge is placed at the center of an uncharged spherical consucting shell 20cm in radius. A) What is the surface charge density on the outer surface of the shell? B) What is the electric field strength at the shell's outer surface?


Homework Equations


A) Charge Density=surface charge/surface area.
B) E=q/(4*pi*E*r^2)

I am unsure if surface charge is the same as the "E" calculated in Part 'B". If it isn't, how do I calculate Surface charge.

The Attempt at a Solution


A) 107330c/m^2 (Doesn't seem correct)
B) results are 53950n/C

Thank you.
Chip
 
Physics news on Phys.org
Surface charge is simply the collective charge on the surface of the sphere (meaning q). The electric field around the surface is E = surface charge density/permittivity of free space.
 
chipperh said:

Homework Statement


a 240nC point charge is placed at the center of an uncharged spherical consucting shell 20cm in radius. A) What is the surface charge density on the outer surface of the shell? B) What is the electric field strength at the shell's outer surface?


Homework Equations


A) Charge Density=surface charge/surface area.
B) E=q/(4*pi*E*r^2)

I am unsure if surface charge is the same as the "E" calculated in Part 'B". If it isn't, how do I calculate Surface charge.

The Attempt at a Solution


A) 107330c/m^2 (Doesn't seem correct)
B) results are 53950n/C

Thank you.
Chip

Because the spherical shell is conducting, in the steady state the electric field in the interior of the shell must be zero. That means that there will be a sufficient surface charge on the interior of the shell to cancel the point charge [lace in the center. Since the conducting shell is neutral, there will be an equal and (opposite sign) surface charge on the outside surface of the shell. The magnitude of these surface charges will have something to do with the magnitude of the charge at the center. If you think about it a bit, you'll also be able to quickly figure out the magnitude of the electric field at the outer surface of the conducting shell.
 
Ahhhh! Thank you. I appreciate the guidance in the correct direction.

My cognitive friction is a result of taking one or two college classes at a time, hence, many classes that are taken during the same semester for full time students are spread out by a few years for me. I have to recall or re-study certain principles, rules and/or methods.

Chip
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top