A point charge Q = +10 nC is placed at the center of an insulating spherical shell of inner radius a = 6 cm and outer radius b = 8 cm. They are contained inside a concentric conducting spherical shell of inner radius c = 10 and outer radius d = 13. The insulator carries a uniform charge density of-12 micro- Coulomb per cubic meter and the conductor .was initially charged with 20 nC

Let's begin by figuring out all the charges and charge densities. What is the net charge of the

)1) نقطة( * insulatorQinsultor

To calculate the net charge of the insulating spherical shell (denoted as \( Q_{\text{insulator}} \)), we will follow these steps:

### Given:
- Charge density of the insulator \( \rho = -12 \, \mu \text{C/m}^3 = -12 \times 10^{-6} \, \text{C/m}^3 \)
- The insulator has inner radius \( a = 6 \, \text{cm} = 0.06 \, \text{m} \)
- The insulator has outer radius \( b = 8 \, \text{cm} = 0.08 \, \text{m} \)

### Step 1: Volume of the insulator

The insulator is a spherical shell, so its volume \( V_{\text{insulator}} \) is the difference between the volumes of two spheres (the outer and inner spheres):

\[
V_{\text{insulator}} = \frac{4}{3} \pi \left( b^3 - a^3 \right)
\]

Substitute the values of \( a \) and \( b \):

\[
V_{\text{insulator}} = \frac{4}{3} \pi \left( (0.08)^3 - (0.06)^3 \right)
\]
\[
V_{\text{insulator}} = \frac{4}{3} \pi \left( 0.000512 - 0.000216 \right)
\]
\[
V_{\text{insulator}} = \frac{4}{3} \pi \times 0.000296
\]
\[
V_{\text{insulator}} \approx 1.24 \times 10^{-3} \, \text{m}^3
\]

### Step 2: Net charge of the insulator

The net charge of the insulator \( Q_{\text{insulator}} \) is the product of the charge density \( \rho \) and the volume \( V_{\text{insulator}} \):

\[
Q_{\text{insulator}} = \rho \times V_{\text{insulator}}
\]
\[
Q_{\text{insulator}} = -12 \times 10^{-6} \, \text{C/m}^3 \times 1.24 \times 10^{-3} \, \text{m}^3
\]
\[
Q_{\text{insulator}} = -1.488 \times 10^{-8} \, \text{C}
\]

Convert to nanocoulombs (nC):

\[
Q_{\text{insulator}} = -14.88 \, \text{nC}
\]

### Final Answer:
The net charge of the insulator \( Q_{\text{insulator}} \) is approximately:

\[
Q_{\text{insulator}} \approx -14.88 \, \text{nC}
\]

Would you like further clarification or details on this calculation?

### 5 Related Questions:
1. What is the electric field inside the insulating shell at a distance \( r = 7 \, \text{cm} \)?
2. How does the net charge on the conducting shell distribute between its inner and outer surfaces?
3. What is the electric field outside the entire system at a distance \( r = 14 \, \text{cm} \)?
4. What is the potential difference between the center of the system and a point outside the conducting shell?
5. How does the presence of the insulator affect the electric field inside the conducting shell?

### Tip:
Remember that spherical symmetry allows simplifications in electrostatics problems, especially when dealing with spherical charge distributions. Use Gauss's law to exploit this symmetry.

A point charge Q = +10 nC is placed at the center of an insulating spherical shell of inner radius a = 6 cm and outer radius b = 8 cm. They are contained inside a concentric conducting spherical shell of inner radius c = 10 cm and outer radius d = 13 cm. The insulator carries a uniform charge density of -12 micro-Coulomb per cubic meter, and the conductor was initially charged with 20 nC. What is the net charge of the insulator?

Learn how to calculate the net charge of an insulating spherical shell with a uniform charge density of -12 micro-Coulombs per cubic meter. This example involves electrostatic concepts, including the volume of a spherical shell and charge density, and applies Gauss's Law. Suitable for undergraduate physics students.

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