At which distance along the centre axis of a uniformaly charged plastic disc of radius R is the magnitude of the electric field equal to one-half the magnitude of the field at the centre of the surface of the disc ?

Options

A. `(R)/(sqrt2)`
B. `(R)/(sqrt3)`
C. `sqrt2R`
D. `sqrt(3R)`

Answer

To solve the problem, we need to find the distance \( x \) along the center axis of a uniformly charged plastic disc of radius \( R \) where the electric field is equal to one-half the magnitude of the electric field at the center of the disc.

### Step-by-Step Solution:

1. **Understanding the Electric Field at the Center of the Disc:**
The electric field \( E_C \) at the center of a uniformly charged disc is given by the formula:
\[ E_C = 2k \pi \sigma \]
where \( k \) is Coulomb's constant and \( \sigma \) is the surface charge density.

2. **Electric Field at a Distance \( x \) Along the Axis:**
The electric field \( E_x \) at a distance \( x \) along the axis of the disc is given by:
\[ E_x = 2k \pi \sigma \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \]

3. **Setting Up the Equation:**
According to the problem, we need to find the distance \( x \) such that:
\[ E_x = \frac{1}{2} E_C \]
Substituting the expression for \( E_C \):
\[ E_x = \frac{1}{2} (2k \pi \sigma) = k \pi \sigma \]

4. **Equating the Two Expressions:**
Now we set \( E_x \) equal to \( k \pi \sigma \):
\[ 2k \pi \sigma \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) = k \pi \sigma \]

5. **Simplifying the Equation:**
Dividing both sides by \( k \pi \sigma \) (assuming \( \sigma \neq 0 \)):
\[ 2 \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) = 1 \]
Simplifying further:
\[ 2 - 2\frac{x}{\sqrt{x^2 + R^2}} = 1 \]
\[ 2\frac{x}{\sqrt{x^2 + R^2}} = 1 \]
\[ \frac{x}{\sqrt{x^2 + R^2}} = \frac{1}{2} \]

6. **Squaring Both Sides:**
Squaring both sides to eliminate the square root:
\[ x^2 = \frac{1}{4}(x^2 + R^2) \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ 4x^2 = x^2 + R^2 \]
\[ 3x^2 = R^2 \]

8. **Solving for \( x \):**
Finally, we solve for \( x \):
\[ x^2 = \frac{R^2}{3} \]
\[ x = \frac{R}{\sqrt{3}} \]

### Final Answer:
The distance \( x \) along the center axis of the uniformly charged plastic disc where the electric field is equal to one-half the magnitude of the field at the center of the disc is:
\[ x = \frac{R}{\sqrt{3}} \]