Simple Electric field question -- Charged sphere

[Rijad Hadzic]

Homework Statement

A sphere with a charge of -3.5 x 10^-9 C and a radius of 1.0x10^-2 m is located at the origin of a coordinate system.

What is the electric field 1.75 x 10^-2 m away from the center of the sphere along the positive y axis?

Homework Equations

e = kQ/r^2

The Attempt at a Solution

Ok so I plug in k and q

(8.99x10^9)(-3.5x10^-9) / r^2

Now for r^2, would the electric field strength be the distance of the coordinate in question, minus the radius of the sphere? I thought the electric field starts from the surface of the sphere?

My book plugged in value (1.75x10^-2) for the value of r, to reach their answer. But on first impression I thought I'm suppose to plug in value of .0075 for r.

Is this a mistake in my understanding or is the books answer wrong?

 

[haruspex]

I thought the electric field starts from the surface of the sphere?

If you were to replace the charge on the sphere with an equal point charge at its centre, would the flux just outside the sphere look any different?

 

[Rijad Hadzic]

If you were to replace the charge on the sphere with an equal point charge at its centre, would the flux just outside the sphere look any different?

Hmm I mean to answer the question, I guess not. Its just kinda hard to grasp at how we know the Electric field starts at the centre..

 

[Rijad Hadzic]

I found this info through google..

"Electric Fields Inside of Charged Conductors. ... If the electrons within a conductorhave assumed an equilibrium state, then the net force upon those electrons is zero. The electric field lines either begin or end upon a charge and in the case of a conductor, the chargeexists solely upon its outer surface."

Is the reason my logic doesn't apply to this case because it doesn't state that the sphere is a conductor?

Because the quoted paragraph says it does start at the surface..

And why would they include the radius of the sphere in this problem then? Do you think it was to throw off people who had my train of thought for this problem?

 

[haruspex]

why would they include the radius of the sphere in this problem then?

Because it is important that it is less than the distance to the point where the field is to be calculated.

 

[Rijad Hadzic]

Because it is important that it is less than the distance to the point where the field is to be calculated.

Okay that makes sense, gotcha! So the field does start at the surface then, right? But because it is so small we can ignore it and say it starts at the centre of the sphere?

 

[haruspex]

the field does start at the surface then, right

Yes.

because it is so small we can ignore it

It only matters that the point of interest is outside the sphere. The absolute size is irrelevant.

 

[Rijad Hadzic]

If you were to replace the charge on the sphere with an equal point charge at its centre, would the flux just outside the sphere look any different?

Wait the answer is the flux just outside of the sphere WOULD look different because it has a larger surface area.. right??

Because it is important that it is less than the distance to the point where the field is to be calculated.

And here your saying the point is basically not inside the sphere, right??

Anyways, so the field starts at the surface.

Wouldn't the start of the field be +radius then? Meaning the distance from the surface and the point doesn't include radius?

Wouldnt I have to use from the surface of the sphere to point p as my r in the electric field strength equation, not the distance from the centre of the sphere to that point?

 

[gneill]

Wait the answer is the flux just outside of the sphere WOULD look different because it has a larger surface area.. right??

Nope. Newton's shell theorem for gravity has its analog for electric fields, and states that a spherically symmetric distribution of charge affects external charges as though all of its charge were concentrated at a point at its center. So at any point outside of the spherical charge distribution the field will be indistinguishable from that of a single point charge of the same total value located at the center of the distribution.

 

[haruspex]

Wait the answer is the flux just outside of the sphere WOULD look different because it has a larger surface area.. right??

The field at a point outside the sphere depends only on the charge on the sphere and the distance from the centre of the sphere to the point.
If you keep the point fixed but vary the radius of the sphere (but keeping the fixed point outside it) the charge density at the surface of the sphere varies inversely to the radius squared since it is spread over a different area. The field just outside the sphere varies correspondingly.
Imagine starting with a point charge and some lines of flux radiating from it. If you expand the point into a sphere, keeping the centre fixed, it absorbs the parts of the flux lines that now fall inside the sphere, wiping those out, but the portions outside the sphere don't change.

For any point inside the sphere there is no field.