# Two conducting hollow spheres, a point charge and electric field prob

• bocobuff
In summary: The net charge would then be the sum of these two contributions. In summary, the problem involves a point charge at the center of a conducting hollow sphere, within another conducting hollow sphere. The net charges for the inner and outer spheres are provided, as well as the distance from the center to point A. The task is to calculate the magnitude of the electric field at point A. Using Gauss' Law, the net flux through a closed surface can be found, which can then be used to determine the net electric field. The net charge will be the sum of the positive and negative contributions from the point charge and the inner sphere.
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## Homework Statement

A point charge is at the center of a conducting hollow sphere, with radius 0.011m, that is within another conducting hollow sphere of radius 0.041m. The point charge is Q0=+4.30e-6 C, the inner sphere has a net charge of Q1=-1.70e-6 C, and the outer sphere has a net charge of Q2=+6.50e-6 C.
Calculate the magnitude of the electric field at a point A located 0.021m from the center.

## Homework Equations

Gauss' Law $$\Phi$$= Qin/ $$\epsilon$$ = E*Areas
Areas=4$$\pi$$r2
$$\epsilon$$=permittivity constant

## The Attempt at a Solution

I've been stuck on this for awhile and don't really know where to begin. I know point A is between the two hollow spheres, one of which is positive, the other negative. But I don't know what to think of the conducting spheres when they aren't in electrostatic equil because they don't have net Q=0. Will the E of point A just be the sum of the electric fields of the spheres and point charge?
EA= E0+E1+E2?

Last edited:
Think about the implications of Gauss Law.

The total flux through a closed surface is going to be the net charge within.

So at P the field you are looking at will be from the point charge and the inner sphere. The outer sphere since you are inside it ... what does that contribute?

It won't contribute anything.

So I tried KQ0/(distance to A)2 - KQ1/(distance to A)2 and it wasn't right.

What's up?

yes it would. you have a positive flux due the point charge and a negative flux due to the inner center. if you can find the net flux, you can find the net electric field. (also, think of A as being an invisible circle instead of a point.)

shelbz23 said:
yes it would. you have a positive flux due the point charge and a negative flux due to the inner center. if you can find the net flux, you can find the net electric field. (also, think of A as being an invisible circle instead of a point.)

alright i got it all figured out. s the net charge only the magnitudes or would you subtract the negative charge?

You would get a positive contribution from Q0 and a negative contribution from Q1.

## 1. What is the purpose of using two conducting hollow spheres in this experiment?

The two conducting hollow spheres are used to create a uniform electric field between them. This allows for accurate measurement and analysis of the electric field caused by the point charge.

## 2. How does the point charge affect the electric field between the two conducting spheres?

The point charge acts as a source of electric field, creating a radial electric field between the two spheres. The strength of the electric field is directly proportional to the magnitude of the point charge.

## 3. How can the electric field between the two spheres be measured?

The electric field can be measured using a voltmeter placed at various points between the two spheres. The voltage measured at each point can be used to calculate the electric field strength.

## 4. How does the distance between the two spheres affect the electric field?

The electric field strength between the two spheres is inversely proportional to the distance between them. This means that as the distance between the spheres increases, the electric field strength decreases.

## 5. What is the significance of using conducting spheres in this experiment?

Using conducting spheres ensures that the electric field remains constant and uniform between the two spheres. This allows for more accurate analysis and measurements of the electric field caused by the point charge.

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