A spherical, non-conducting shell of inner radius r1 = 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC distributed uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 12 cm from the center of the shell? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

The electric field is the force acting on a charge located in a region in space. The electric field at a distance r from the center of the shell is 2.8x10⁶N/C

What is the electric field?

The electric field (E) can be defined as a region in the space that interacts with electric charges or charged bodies through electric forces.

It is a vectorial field in which an electric charge (q) suffers the effect of the electric force (F). In other words, it represents the force that is acting on a charge located in a certain region.

E = F/q

It is expressed in newton/coulomb (N/C)

In the exposed example, we have the following data,

spherical, non-conducting shell
inner radius r1 = 10 cm
outer radius r2 = 15 cm
charge Q = 15 μC uniformly throughout the volume of the shell
distance r = 12 cm from the center of the shell
k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²

We need to get the value of the electric field at the given distance r.

E = Q/4πεr² = kQ/r²

Since the r value is located between r1 and r2, we need to get the electric field at that intermedia distance, r. This is a new ratio.

The total charge Q is distributed in the whole area of the shell thickness or density, delimited by r1 and r2.

But to get the electric field at the intermedia distance, r, we only need to get the charge q lower than Q and distributed between the r and r1, which represents a volume v lower than the total volume V of the shell.

So to get the q value, we need to consider the charge volumetric density, σ.

σ = Q/V

Where

Q is the total charge
V is the total volume

But to get Q/V, we need to get the total shell volume value, V.

V = 4/3 π r³

V = 4/3 π (r2³ - r1³)

V = 4/3 π (0.15³ - 0.1³)

V = 4/3 π (0.00237)

V = 0.00993m³

Now that we have the total volume, and we know the value of the total charge, we need to get the density of the shell.

σ = Q/V

σ = 15 μC / 0.00993m³

σ = 15x10⁻⁶ C / 0.00993m³

σ = 0.00151 C/m³

Now, we will calculate the volume v between r and r1, which is what we are interested in.

v = 4/3 π r³

v = 4/3 π (r³ - r1³)

v = 4/3 π (0.12³ - 0.1³)

v = 4/3 π (0.0007)

v = 0.00305m³

Finally, using this new v value and the σ value, we can get the charge q distributed throughout the volume v.

q = vσ

q = 0.00305m³ x 0.00151 C/m³

q = 4.6x10⁻⁶ C

Now we can get the electric field at a distance r from the center of the shell

E = kq/r²

E = ((8.99 × 10⁹ N ∙ m²/C²) x (4.6x10⁻⁶ C) ) / 0.12²m

E = 2.8x10⁶N/C

Hence, the electric field at a distance r is 2.8x10⁶N/C

You can learn more about electric field at

https://brainly.com/question/8971780