Consider an insulating sphere with 10 microCoulombs of charge uniformly distributed through its volume. The sphere is surrounded by a conducting spherical shell that has a total charge of -3 microCoulombs. Outside the conucting shell is an insulating shell with total charge 20 microCoulombs uniformly distributed through its volume. In the figure below, which shows a cross section of the structure, the darker shaded regions are insulators and the lighter shaded region is the condcutor.

What is the magnitude of the electric field at point A halfway between the outer surface of the insulatng sphere and the inner surface of the conducting shell, if that point is 5 centimeters from the center of the sphere? Give the field magnitude in Newtons/Coulomb.

To find the magnitude of the electric field at point A, we can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).

In this case, we can imagine a sphere with radius 5 cm centered at the middle of the region between the outer surface of the insulating sphere and the inner surface of the conducting shell. Since the charge distribution is symmetric, the electric field at point A will be the same for any direction we choose.

The charge enclosed within this Gaussian surface is the sum of the charge of the insulating sphere (10 μC) and the charge of the insulating shell (20 μC), which is a total of 30 μC.

The permittivity of free space, ε₀, is a constant equal to approximately 8.85 x 10⁻¹² C²/N·m².

We can now use Gauss's Law to calculate the electric field magnitude at point A:

Electric flux = Electric field * surface area

The electric flux is equal to the charge enclosed divided by ε₀:

Electric flux = (30 μC) / (8.85 x 10⁻¹² C²/N·m²)

Now we can rearrange the equation to solve for the electric field:

Electric field = Electric flux / surface area

The surface area of the Gaussian surface is 4π(0.05 m)², as it is a sphere with radius 5 cm.

Plugging in the values:

Electric field = (30 μC) / (8.85 x 10⁻¹² C²/N·m²) / (4π(0.05 m)²)

Simplifying the calculation:

Electric field ≈ 2.69 x 10⁷ N/C

Therefore, the magnitude of the electric field at point A is approximately 2.69 x 10⁷ N/C.

To find the magnitude of the electric field at point A, we can use the principle of superposition. The electric field at point A will be the sum of the electric fields due to the individual charges at the given positions.

Step 1: Electric Field due to the Insulating Sphere
The electric field at point A due to the insulating sphere can be found using Coulomb's Law. Coulomb's Law states that the electric field due to a point charge is given by the equation:

E = k * (q / r^2)

where E is the electric field, k is Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.

In this case, we have a uniformly distributed charge throughout the volume of the insulating sphere. To find the electric field due to the entire sphere, we can consider a small charge element within the sphere and integrate over the entire volume.

However, since the sphere is insulating, the electric field inside the sphere is zero due to the repulsion of charges.

Step 2: Electric Field due to the Conducting Shell
The electric field inside a conductor in electrostatic equilibrium is also zero. This means that the charges residing on the inner surface of the conducting shell will redistribute themselves such that the electric field inside cancels out.

Step 3: Electric Field due to the Insulating Shell
We can again use Coulomb's Law to find the electric field due to the insulating shell at point A. Since the total charge of the insulating shell is uniformly distributed, the electric field due to the entire shell can be found by considering a small charge element within the shell and integrating over the volume.

The result of the calculation will give us the electric field at point A due to only the insulating shell.

Step 4: Combine the Electric Fields
Since the electric field is a vector quantity, we need to sum the electric fields calculated from steps 1, 2, and 3 to find the total electric field at point A.

Finally, we can express the magnitude of the electric field at point A as a single value in Newtons/Coulomb, using the information provided.