Two concentric conducting spherical shells produce a radially outward electric field of magnitude 49,000 N/C at a point 4.10 m from the center of the shells. The outer surface of the larger shell has a radius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of charge on the outer surface of the larger shell. (k = 1/4ε0 = 8.99 × 109 N · m2/C2)

Magnitude of the electric field just outside the outer surface is:

E = 49,000 N/C (4.10/3.75 )^2

equate this to sigma/epsilon_0 to find the surface charge density.

To solve this problem, we can use Gauss's law to find the electric field at a distance r inside the inner shell:

ε0 * E = Qenc / ε0

Where ε0 is the permittivity of free space, E is the electric field magnitude, and Qenc is the charge enclosed within a surface. Since the inner shell is conducting, the electric field within it will be zero. Therefore, the electric field at a distance r inside the inner shell is the same as the electric field at the surface of the inner shell.

We know that the electric field at a distance of 4.10 m from the center of the shells is 49,000 N/C. Using this information, we can find the electric field at the inner surface of the larger shell by using the equation:

ε0 * E = Qenc / ε0

Rearranging this equation, we get:

Qenc = ε0 * E

Qenc = (8.99 × 10^9 N · m^2/C^2) * (4.10 m)

Qenc = 3.68 × 10^10 C

Now, let's find the charge on the outer surface of the larger shell. Since the inner shell contains a charge of -5.30 µC, and the charge enclosed within the larger shell is the sum of the charges on the inner and outer surfaces, we can calculate the charge on the outer surface:

Qouter = Qenc + (-5.30 × 10^-6 C)

Qouter = 3.68 × 10^10 C + (-5.30 × 10^-6 C)

Qouter = 3.68 × 10^10 C - 5.30 × 10^-6 C

Qouter = 3.68 × 10^10 C - 0.0000053 C

Qouter = 3.67999994 × 10^10 C

Therefore, the amount of charge on the outer surface of the larger shell is approximately 3.68 × 10^10 C.

To find the charge on the outer surface of the larger shell, we need to first find the electric field due to the charge on the inner shell and then use Gauss's law to relate the electric field to the charge.

1. Find the electric field due to the inner shell:
Gauss's law states that the electric field inside a Gaussian surface is proportional to the charge enclosed by the surface. Since the electric field is radially outward, we can use a Gaussian surface in the form of a spherical shell centered at the center of the shells.

The electric field due to the inner shell only depends on the charge enclosed within this Gaussian surface, which is -5.30 μC.

Using Gauss's law, we can write:
E * 4πr^2 = q / ε0
where E is the electric field at a distance r from the center, q is the charge enclosed within the Gaussian surface, and ε0 is the vacuum permittivity constant.

Rearranging the equation, we get:
E = (q / ε0) / (4πr^2)

Plug in the values:
E = (-5.30 × 10^-6 C) / (8.99 × 10^9 N · m^2/C^2) / (4π(4.10 m)^2)
E = -2.72 × 10^6 N/C

2. Find the electric field due to the outer shell:
Since the electric field is radially outward, the electric field due to the outer shell is simply 49,000 N/C at the given point.

3. Find the total electric field at the given point:
The total electric field at the given point is the vector sum of the electric fields due to the inner and outer shells.

E_total = E_inner + E_outer
E_total = -2.72 × 10^6 N/C + 49,000 N/C
E_total = -2.67 × 10^6 N/C

4. Use Gauss's law to relate the electric field to the charge on the outer surface of the larger shell:
The electric field outside the larger shell only depends on the charge on the outer surface. We can use a Gaussian surface outside the shell.

Using Gauss's law, we get:
E * 4πr^2 = q / ε0

Rearranging the equation, we get:
q = E * 4πr^2 * ε0

Plug in the values:
q = (-2.67 × 10^6 N/C) * 4π(3.75 m)^2 * (8.99 × 10^9 N · m^2/C^2)
q ≈ 1.97 × 10^-6 C

So, the amount of charge on the outer surface of the larger shell is approximately 1.97 μC.

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