Electric field at the centre of uniformly charge hemispherical shell of surface charge density `sigma` is `(sigma)/(n epsi_(0))` then find the value of `n`.

Answer

To find the value of \( n \) in the expression for the electric field at the center of a uniformly charged hemispherical shell, we can follow these steps:

### Step 1: Understand the Problem
We are given a uniformly charged hemispherical shell with a surface charge density \( \sigma \). The electric field at the center of this shell is given as \( \frac{\sigma}{n \epsilon_0} \). Our goal is to find the value of \( n \).

### Step 2: Consider an Elementary Ring
Consider an elementary ring at an angle \( \theta \) from the base of the hemisphere. The thickness of this ring is \( R \, d\theta \), where \( R \) is the radius of the hemisphere.

### Step 3: Determine the Radius of the Ring
The radius of the ring can be expressed as:
\[ r = R \sin \theta \]
The distance from the center of the ring to the center of the hemisphere is:
\[ x = R \cos \theta \]

### Step 4: Calculate the Charge on the Elementary Ring
The charge \( dQ \) on the elementary ring can be calculated as:
\[ dQ = \sigma \, dA \]
where \( dA = 2\pi r \, dr = 2\pi (R \sin \theta)(R \, d\theta) = 2\pi R^2 \sin \theta \, d\theta \).

Thus, we have:
\[ dQ = \sigma \cdot 2\pi R^2 \sin \theta \, d\theta \]

### Step 5: Calculate the Electric Field Contribution from the Ring
The electric field \( dE \) at the center due to the elementary ring is given by:
\[ dE = k \frac{dQ \cdot x}{(R^2 \sin^2 \theta + x^2)^{3/2}} \]
Substituting \( dQ \) and \( x \):
\[ dE = k \frac{\sigma \cdot 2\pi R^2 \sin \theta \, d\theta \cdot (R \cos \theta)}{(R^2 \sin^2 \theta + R^2 \cos^2 \theta)^{3/2}} \]

### Step 6: Simplify the Expression
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ dE = k \frac{\sigma \cdot 2\pi R^2 \sin \theta \, d\theta \cdot (R \cos \theta)}{(R^2)^{3/2}} = k \frac{\sigma \cdot 2\pi R^2 \sin \theta \, d\theta \cdot (R \cos \theta)}{R^3} \]
This simplifies to:
\[ dE = k \frac{\sigma \cdot 2\pi \sin \theta \cos \theta}{R} \, d\theta \]

### Step 7: Integrate to Find Total Electric Field
The total electric field \( E \) at the center is obtained by integrating \( dE \) from \( 0 \) to \( \frac{\pi}{2} \):
\[ E = \int_0^{\frac{\pi}{2}} dE = k \frac{\sigma}{R} \int_0^{\frac{\pi}{2}} 2\pi \sin \theta \cos \theta \, d\theta \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ E = k \frac{\sigma}{R} \int_0^{\frac{\pi}{2}} \sin 2\theta \, d\theta \]

### Step 8: Evaluate the Integral
The integral of \( \sin 2\theta \) is:
\[ -\frac{1}{2} \cos 2\theta \bigg|_0^{\frac{\pi}{2}} = -\frac{1}{2} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{2} \left( -1 - 1 \right) = 1 \]
Thus:
\[ E = k \frac{\sigma}{R} \cdot 2\pi \cdot 1 = k \frac{\sigma \cdot 2\pi}{R} \]

### Step 9: Relate to the Given Expression
We know that:
\[ E = \frac{\sigma}{n \epsilon_0} \]
Using \( k = \frac{1}{4\pi \epsilon_0} \):
\[ E = \frac{\sigma}{4\pi \epsilon_0 R} \]

### Step 10: Compare and Solve for \( n \)
Comparing both expressions for \( E \):
\[ \frac{\sigma}{4\pi \epsilon_0 R} = \frac{\sigma}{n \epsilon_0} \]
Thus:
\[ n = 4\pi R \]
Since we are looking for the numerical value of \( n \) which is independent of \( R \), we find that \( n = 4 \).

### Final Answer
The value of \( n \) is \( 4 \).

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