A charged disc of radius 3 m has charge density `sigma`. Electric field at a distance of 4 m from its centre on axis is _______

Options

A. `(3sigma)/(5epsilon_(0))`
B. `(sigma)/(10 epsilon_(0))`
C. `(sigma)/(8epsilon_(0))`
D. None

Answer

To find the electric field at a distance of 4 m from the center of a charged disc with a radius of 3 m and charge density \( \sigma \), we can use the formula for the electric field along the axis of a uniformly charged disc. The formula is given by:

\[ E = \frac{\sigma}{2 \epsilon_0} \left( 1 - \frac{x}{\sqrt{r^2 + x^2}} \right) \]

where:
- \( E \) is the electric field,
- \( \sigma \) is the surface charge density,
- \( \epsilon_0 \) is the permittivity of free space,
- \( r \) is the radius of the disc,
- \( x \) is the distance from the center of the disc along the axis.

### Step 1: Identify the given values
- Radius of the disc, \( r = 3 \, \text{m} \)
- Distance from the center, \( x = 4 \, \text{m} \)

### Step 2: Substitute the values into the formula
We substitute \( r \) and \( x \) into the electric field formula:

\[ E = \frac{\sigma}{2 \epsilon_0} \left( 1 - \frac{4}{\sqrt{3^2 + 4^2}} \right) \]

### Step 3: Calculate \( r^2 + x^2 \)
Calculate \( r^2 + x^2 \):

\[ r^2 + x^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

### Step 4: Find the square root
Now, find the square root of \( 25 \):

\[ \sqrt{25} = 5 \]

### Step 5: Substitute back into the electric field formula
Now substitute back into the formula:

\[ E = \frac{\sigma}{2 \epsilon_0} \left( 1 - \frac{4}{5} \right) \]

### Step 6: Simplify the expression
Now simplify the expression inside the parentheses:

\[ 1 - \frac{4}{5} = \frac{1}{5} \]

So, we have:

\[ E = \frac{\sigma}{2 \epsilon_0} \cdot \frac{1}{5} \]

### Step 7: Final simplification
This can be simplified to:

\[ E = \frac{\sigma}{10 \epsilon_0} \]

### Conclusion
Thus, the electric field at a distance of 4 m from the center of the charged disc is:

\[ E = \frac{\sigma}{10 \epsilon_0} \]

The correct answer is **\( \frac{\sigma}{10 \epsilon_0} \)**.

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