You are Here: Home >< Physics

Conducting Sphere surface charge density Watch

Announcements
1. A hollow metal sphere has 7 cm and 10 cm inner and outer radii, respectively. The surface charge density on the inside surface is −300nC/m2. The surface charge density on the exterior surface is +300nC/m2What is the electric field 4cm from the center and 12cm from the center?

The overall solution to this problem seems like it should be simple, however the one thing confusing me is what I should do to convert surface charge density to charge for use in formulas.
Lambda=Q/L so Q= Lambda(L). What value should I use for L? Should it be the distance (4cm for the first question and 12cm for the second) or should it be the inner/outer radius of the actual sphere?
2. (Original post by PatchworkTeapot)
A hollow metal sphere has 7 cm and 10 cm inner and outer radii, respectively. The surface charge density on the inside surface is −300nC/m2. The surface charge density on the exterior surface is +300nC/m2What is the electric field 4cm from the center and 12cm from the center?

The overall solution to this problem seems like it should be simple, however the one thing confusing me is what I should do to convert surface charge density to charge for use in formulas.
Lambda=Q/L so Q= Lambda(L). What value should I use for L? Should it be the distance (4cm for the first question and 12cm for the second) or should it be the inner/outer radius of the actual sphere?
multiply by the area.

Give it a try now
3. (Original post by Daniel Atieh)
multiply by the area.

Give it a try now
Do you mean multiply the surface charge density by the area of the actual sphere or by an area of 4cm radius?
4. Am I missing something here? 4cm from the centre is inside the hollow, where there will be no field. 12cm from the centre will be outside the sphere, where there will also be no field.
5. (Original post by mik1a)
Am I missing something here? 4cm from the centre is inside the hollow, where there will be no field. 12cm from the centre will be outside the sphere, where there will also be no field.

I thought the inner one would be zero too, but apparently that is not correct. The outer one should be treated like a point charge, I think, but I don't know which value you use to find the charge from the surface charge density
6. Ah, my mistake. The inner one is certainly true because you can construct a Gauss' law surace integral on a spherical shell of radius 4cm that contains no charge, thus the electric field passing through the shell must be zero, therefore by symmetry the field is zero inside the hollow.

Outside it is a different story- although the surface charge densities are the same, the outer spherical surface obviously has a larger surfacre area than the inner surface, so there is more charge on the outer surface than the inner one. So the net charge (the important thing) is the difference between the charges of the two surfaces (i.e. +300 nC times (4/3)*pi*(r2-r1)^2, the surface area differential), and you can again use Gauss's law to calculate the electric field that results.

I'm assuming you're familiar with Gauss's law (integral of perpendicular electric field over a closed surface equals the integral of the charge density in the volume enclosed), and are happy applying it to spherically symmetric systems.
7. For charge to be separated on a conducting object something must be forcing the charges to separate. Since the surface charge density on the inside surface is negative, there must be a positive point charge in the center of the hollow sphere
8. (Original post by theheuman)
For charge to be separated on a conducting object something must be forcing the charges to separate. Since the surface charge density on the inside surface is negative, there must be a positive point charge in the center of the hollow sphere
How would you determine its magnitude for solving the problem?

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 8, 2016
Today on TSR

This is how Oxbridge students do it.

This compliment left me speechless...

Discussions on TSR

• Latest
• See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Discussions on TSR

• Latest
• See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.