Solving Spherical Conductor: Net Charge and Electric Field

[tica86]

A solid spherical conductor of radius 1.50 m has a small spherical cavity of radius 0.50 m with the same center. The net charge of the conductor is zero. An object with a net charge of 4.00 nC is placed inside the cavity in such a way that it is isolated from the conductor. I know the answers I just want to be able to understand the concept.

1)What is the charge on the inner cavity surface?

I know it is -4.00nC but why??


2)What is the electric field magnitude 3.00m from the center of the sphere?
E(4pi*r^2)=Q/E0
THE ANSWER is 3.996 V/m again I don't understand..

 

[Falcons]

For 1) It's induction. The 4 nC charge pulls a minus 4 nC charge towards itself from the conductor. 2) Is Gauss' law, and there really isn't another way to explain it.

 

[SammyS]

You can also explain (1) using Gauss's Law. Consider a sphere of radius 0.60 m (It could anything greater than 0.50 and less than 1.50 m.) with center at the same location as the other two spheres. The electric field is zero, everywhere on the surface of this sphere.

How much flux passes through this closed surface?

What is the net charge enclosed by the surface of this sphere?

 

[gomunkul51]

perfect symmetry -> use Gauss' Law :)

(Gaussian surface inside the conduction outer shell)

 

[rwisz]

I can explain the concept behind solving this problem. Firstly, in a conductor, the charges are free to move and redistribute themselves in order to reach a state of equilibrium. This means that the net charge on the conductor will always be zero, as the charges will redistribute themselves to cancel out any excess charge.

Now, in this specific scenario, we have a solid spherical conductor with a small spherical cavity inside. The net charge on the conductor is zero, meaning that the total charge inside the conductor (including the cavity) must also be zero. This is because the charges inside the conductor will redistribute themselves to cancel out any charge inside the cavity as well.

Therefore, when we place an object with a net charge of 4.00 nC inside the cavity, the charges inside the conductor will redistribute themselves to cancel out this charge. Since the net charge on the conductor must remain zero, the charges on the inner cavity surface must be equal in magnitude but opposite in sign to the charge on the object (in this case, -4.00 nC).

Now, for the second question, we need to use the formula for electric field intensity (E) at a distance (r) from a point charge (Q), which is given by E=Q/(4πε0r^2), where ε0 is the permittivity of free space. In this case, the point charge is the object with a net charge of 4.00 nC inside the cavity.

So, when we plug in the values, we get E=(4.00 nC)/(4πε0(3.00 m)^2) = 3.996 V/m. This is the electric field magnitude at a distance of 3.00 m from the center of the sphere, which is the same as the distance from the center of the cavity. This is because the charges inside the conductor will redistribute themselves to cancel out the charge on the object, creating a uniform electric field throughout the conductor.

I hope this helps to clarify the concept behind solving this problem. Let me know if you have any further questions.