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.

3.6*10^7 N/C

To determine the electric field at point A, we can make use of Gauss's law. According to Gauss's law, the electric field at a point outside a charged spherical shell is equivalent to the field created by a point charge located at the center of the sphere. In this case, the electric field at point A will be due to the charge on the insulating sphere.

We can calculate the electric field using the formula,

E = (k * Q) / r^2

Where:
E is the electric field in Newtons/Coulomb,
k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2),
Q is the charge enclosed within the Gaussian surface, and
r is the distance from the point charge to the field point.

In this case, we have a uniformly distributed charge symmetrically distributed within the insulating sphere. The magnitude of the charge enclosed within the Gaussian surface will be the ratio of the volume of the Gaussian surface to the total volume of the insulating sphere multiplied by the total charge of the insulating sphere.

First, let's calculate the charge enclosed:

Charge enclosed = (Volume of Gaussian surface / Total volume of insulating sphere) * Total charge of insulating sphere

The volume of the Gaussian surface is given by:

Volume of Gaussian surface = (4/3) * π * r^3

The total volume of the insulating sphere is given by:

Total volume of insulating sphere = (4/3) * π * R^3

Where R is the radius of the insulating sphere.

Next, we can calculate the charge enclosed:

Charge enclosed = [(4/3) * π * r^3 / (4/3) * π * R^3] * (10 x 10^-6 C)

Simplifying this expression:

Charge enclosed = (r^3 / R^3) * (10 x 10^-6 C)

Now, we can calculate the electric field at point A using the formula mentioned earlier:

E = (k * Charge enclosed) / r^2

E = (8.99 x 10^9 N m^2/C^2) * [(r^3 / R^3) * (10 x 10^-6 C)] / (0.05 m)^2

Simplifying this expression will give us the magnitude of the electric field at point A in N/C.

To determine the magnitude of the electric field at point A, we need to consider the contributions from each of the charges present in the system. We can break down the problem into smaller parts and analyze the electric field due to each charge separately.

Step 1: Electric Field due to the insulating sphere
Since the charge in the insulating sphere is uniformly distributed throughout its volume, the electric field inside the sphere is zero. Thus, the only contribution to the electric field at point A comes from the charges outside the insulating sphere.

Step 2: Electric Field due to the conducting shell
Inside a conductor, the electric field is always zero. Therefore, the presence of the conducting shell does not contribute to the electric field at point A.

Step 3: Electric Field due to the insulating shell
The insulating shell has a total charge of 20 microCoulombs uniformly distributed throughout its volume. The electric field due to a uniformly charged insulating sphere can be calculated using the formula for the electric field of a point charge outside the sphere.

To calculate the electric field contribution from the insulating shell at point A, we consider the point charges located at various distances from point A within the shell. Since the insulating shell is spherical, the electric field contribution from each point charge will be in the same direction at point A.

To calculate the electric field at point A due to the entire insulating shell, we need to integrate the electric field contributions from all the infinitesimally small point charges within the shell. The formula for the electric field due to a point charge is:

E = k * (q / r^2)

where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point where the electric field is being measured.

Integrating the electric field contributions from all the infinitesimally small point charges within the shell gives the final electric field at point A.

Step 4: Determine the distance between the charges and point A
The problem states that point A is halfway between the outer surface of the insulating sphere and the inner surface of the conducting shell and is 5 centimeters from the center of the sphere. This means that the distance between the charges within the insulating shell and point A is also 5 centimeters.

Step 5: Perform the calculations
To calculate the electric field at point A, we need to determine the total charge within the insulating shell and then integrate the electric field contribution over all the charges within the shell.

The total charge within the insulating shell is given as 20 microCoulombs. We can then calculate the electric field at point A using the formula mentioned earlier:

E = k * (q / r^2)

where
k = 9 x 10^9 Nm^2/C^2 (electrostatic constant)
q = 20 x 10^-6 C (charge within the insulating shell)
r = 0.05 m (distance between the charges within the insulating shell and point A)

Plugging the values into the formula, we can calculate the electric field at point A in Newtons/Coulomb.