The magnitude of electric field `vecE` in the annular region of a charged cylindrical capacitor.

Options

A. is the same throughout
B. is higher near the outer cylinder than near the inner cylinder
C. varies as `(1)/(r^(2))` where r is the distance from the axis
D. varies as `(1)/(r^(3))` where r is the distance from the axis.

Answer

To find the magnitude of the electric field \( \vec{E} \) in the annular region of a charged cylindrical capacitor, we can follow these steps:

### Step 1: Understand the Configuration
We have a cylindrical capacitor consisting of two concentric cylinders. The inner cylinder has a linear charge density \( \lambda \) and the outer cylinder is neutral or grounded. We will analyze the electric field in the annular region between the two cylinders.

### Step 2: Apply Gauss's Law
To find the electric field, we can use Gauss's Law, which states:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux through a closed surface, \( Q_{\text{enc}} \) is the charge enclosed by that surface, and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Choose a Gaussian Surface
We consider a cylindrical Gaussian surface of radius \( r \) (where \( r \) is greater than the radius of the inner cylinder) and length \( L \). The electric field \( \vec{E} \) is uniform and radial at this surface.

### Step 4: Calculate the Electric Flux
The electric flux through the curved surface of the cylinder is given by:
\[ \Phi_E = E \cdot A = E \cdot (2 \pi r L) \]
where \( A \) is the area of the curved surface of the cylinder.

### Step 5: Determine the Enclosed Charge
The charge enclosed by the Gaussian surface is the charge on the inner cylinder:
\[ Q_{\text{enc}} = \lambda L \]
where \( \lambda \) is the linear charge density.

### Step 6: Apply Gauss's Law
Substituting the expressions for electric flux and enclosed charge into Gauss's Law, we get:
\[ E \cdot (2 \pi r L) = \frac{\lambda L}{\epsilon_0} \]

### Step 7: Solve for the Electric Field
Now, we can solve for \( E \):
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]
This shows that the electric field \( E \) in the annular region is inversely proportional to the distance \( r \) from the axis of the cylinder.

### Final Expression
Thus, the magnitude of the electric field in the annular region of the charged cylindrical capacitor is:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]