A spherical shell 25 cm
 in diameter carries a total charge of 80 μC
 distributed uniformly over its surface. A 5.5-μC
 point charge is located at the center of the shell.What is the electric field strength 7.5  cm
  from the center?

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.

A spherical shell 25 cm in diameter carries a total charge of 80 μC distributed uniformly over its surface. A 5.5-μC point charge is located at the center of the shell. What is the electric field strength 7.5 cm from the center?

Learn how to calculate the electric field strength 7.5 cm from the center of a spherical shell with a total charge of 80 μC and a point charge of 5.5 μC at the center. This example covers electrostatic concepts using Gauss's Law. Suitable for undergraduate physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation