Two spherical shells have a common center. A charge of `-1.6 xx 10 ^(-6)` C is spread uniformaly over the inner shell, which has a radius of `0.050m.` A charge of `+ 5.1 xx 10 ^(-6)C` is spread uniformly over the outer shell, which has a radius of `0.15m.` Find the magnitude and direction of the electric field at distacne (measured from the common center) of `0.20m.`

Options

A. `7.9xx10 ^(5)N//C,` radially outward
B. `5.9 xx 10 ^(5)N//C,`radially inward
C. `3.2xx10^(5) N//C,` radially outward
D. zero N/C

Answer

To find the magnitude and direction of the electric field at a distance of 0.20 m from the common center of two concentric spherical shells, we can use Gauss's Law and the principle of superposition. Here’s a step-by-step solution:

### Step 1: Identify the charges and their distribution
- Inner shell charge, \( Q_1 = -1.6 \times 10^{-6} \, \text{C} \)
- Outer shell charge, \( Q_2 = +5.1 \times 10^{-6} \, \text{C} \)
- Inner shell radius, \( R_1 = 0.050 \, \text{m} \)
- Outer shell radius, \( R_2 = 0.15 \, \text{m} \)

### Step 2: Determine the distance from the center
We need to find the electric field at a distance \( r = 0.20 \, \text{m} \), which is outside both shells (since \( r > R_2 \)).

### Step 3: Apply Gauss's Law
For a point outside both shells, the total charge enclosed by a Gaussian surface at \( r = 0.20 \, \text{m} \) is the sum of the charges on both shells:
\[ Q_{\text{enc}} = Q_1 + Q_2 = -1.6 \times 10^{-6} \, \text{C} + 5.1 \times 10^{-6} \, \text{C} = 3.5 \times 10^{-6} \, \text{C} \]

### Step 4: Calculate the electric field using the formula
According to Gauss's Law, the electric field \( E \) at a distance \( r \) from the center is given by:
\[ E = \frac{k \cdot Q_{\text{enc}}}{r^2} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Substituting the values:
\[ E = \frac{9 \times 10^9 \cdot 3.5 \times 10^{-6}}{(0.20)^2} \]

### Step 5: Calculate \( E \)
Calculating the denominator:
\[ (0.20)^2 = 0.04 \]
Now substituting back into the equation:
\[ E = \frac{9 \times 10^9 \cdot 3.5 \times 10^{-6}}{0.04} \]
\[ E = \frac{31.5 \times 10^3}{0.04} = 7.875 \times 10^5 \, \text{N/C} \]

### Step 6: Determine the direction of the electric field
Since the net charge \( Q_{\text{enc}} \) is positive, the electric field will point radially outward from the center.

### Final Answer
The magnitude of the electric field at a distance of 0.20 m from the common center is approximately \( 7.9 \times 10^5 \, \text{N/C} \) and the direction is radially outward.

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