To solve this, we use Gauss's Law which states that the electric field produced by a spherical charge distribution outside the charged shell is the same as if all the charge were concentrated at the center.

Given:

• Diameter of the spherical shell, $D = 25 \, \text{cm}$ ⇒ Radius of the shell, $R_{\text{shell}} = \frac{25}{2} = 12.5 \, \text{cm}$
• Total charge on the shell, $Q_{\text{shell}} = 80 \, \mu\text{C} = 80 \times 10^{-6} \, \text{C}$
• Charge at the center of the shell, $Q_{\text{center}} = 5.5 \, \mu\text{C} = 5.5 \times 10^{-6} \, \text{C}$
• Distance from the center to the point where we need the electric field, $r = 7.5 \, \text{cm}$

Step 1: Understanding the electric field inside the spherical shell

• Inside the shell, the electric field due to the shell’s charge is zero, according to Gauss's Law. This is because a uniformly charged spherical shell creates no electric field inside it.

• Therefore, the only charge contributing to the electric field at $r = 7.5 \, \text{cm}$ (which is inside the shell) is the point charge at the center.

Step 2: Electric field due to the point charge at the center

The electric field at a distance $r$ from a point charge $Q_{\text{center}}$ is given by:

$E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_{\text{center}}}{r^2}$

Where:

• $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot \text{m}^2$ (permittivity of free space),
• $r = 7.5 \, \text{cm} = 0.075 \, \text{m}$,
• $Q_{\text{center}} = 5.5 \times 10^{-6} \, \text{C}$.

Now, substitute the values into the equation:

$E = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{5.5 \times 10^{-6}}{(0.075)^2}$

Let’s calculate that.The electric field strength at a distance of 7.5 cm from the center of the spherical shell is approximately $8.79 \times 10^6 \, \text{N/C}$.

Would you like more details or have any other questions?

Here are 5 related questions for further exploration:

1. What would the electric field be at a distance greater than the shell's radius?
2. How would the electric field change if the central charge were doubled?
3. What happens to the electric field inside the shell if the charge on the shell is non-uniform?
4. How is the electric potential related to the electric field in this scenario?
5. What would the electric field be at the shell’s surface?

Tip: Remember, a spherical shell with uniform charge distribution creates no electric field inside itself.