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.