# Why is this 0? Help Gauss's law

## Homework Statement

A long straight metal rod has a radius of 5.00cm and a charge per unit length of 30.0nC/m. Find the electric field

a) 3.00cm

## The Attempt at a Solution

It says a) = 0

I did some math and I got

$$2k\frac{\lambda}{r}$$

The rod is solid, so should a E-field be radially outward from its symmetrical axis?

## Answers and Replies

I am assuming that, what you wrote as, "the electric field a) 3.00cm" means "the electric field at 3.00cm from the center of the wire"

Answer to this question first: If you have a conducting sphere, carrying a charge say Q, than what is the electric field at a point inside it??

supratim1
Gold Member
first check theory on electric field inside conductors.

I am assuming that, what you wrote as, "the electric field a) 3.00cm" means "the electric field at 3.00cm from the center of the wire"

Answer to this question first: If you have a conducting sphere, carrying a charge say Q, than what is the electric field at a point inside it??

$$k\frac{q}{a^3}r$$ for a < r

it is a metal rod. so all the charge would reside on the surface only, none can stay in the bulk.

so the field inside the rod is zero.

$$k\frac{q}{a^3}r$$ for a < r

its for symmetrical charge distribution throughout volume. metals cant have that, all of their charge resides on surface!!! Can you tell why?

and i believe it must be "r<a"

it is a metal rod. so all the charge would reside on the surface only, none can stay in the bulk.

so the field inside the rod is zero.

So, but a metal sphere can too? Isn't that what my equation is?

Oh wait, I think I get it. The key here is that my equation is for an insulating sphere, not a conducting one. Hence.

So, but a metal sphere can too? Isn't that what my equation is?

Oh wait, I think I get it. The key here is that my equation is for an insulating sphere, not a conducting one. Hence.

Yes now you're right

Yes now you're right

If this was not a "metal" rod, or "insulating rod", then my numbers in the above would be right, right?

i cant get you

i cant get you

If it is insulating, all the charges would be stuck and won't be all on the surface

$$k\frac{q}{a^3}r$$ for a < r

If you are saying that this formula is for insulating sphere then you are right!!!

And also for the explanation, the charges will not move to the surface!!!

can you find the same for insulating cylinder?

$$k\frac{q}{a^3}r$$ for r<a

If you are saying that this formula is for insulating sphere then you are right!!!

And also for the explanation, the charges will not move to the surface!!!

can you find the same for insulating cylinder?
just draw a gaussian surface with r < radius of cylinder and coaxial with cylinder and use gauss theorm