A uniform electric field E=500 N /C passes through a hemispherical surface of radius R=1.2 mas shown in the figure. The net electric flux (in SI units) through the hemispherical surface only is `N pi`. Find the value of N.
'(##CEN_KSR_PHY_JEE_C18_E01_008_Q01##)'

Answer

To find the value of \( N \) in the expression for the electric flux through a hemispherical surface in a uniform electric field, we can follow these steps:

### Step 1: Understand the Electric Flux Formula
The electric flux \( \Phi_E \) through a surface is given by the formula:
\[ \Phi_E = E \cdot A \cdot \cos(\theta) \]
where:
- \( E \) is the electric field strength,
- \( A \) is the area of the surface,
- \( \theta \) is the angle between the electric field and the normal to the surface.

### Step 2: Identify the Parameters
From the question:
- The electric field \( E = 500 \, \text{N/C} \).
- The radius of the hemispherical surface \( R = 1.2 \, \text{m} \).

### Step 3: Calculate the Area of the Hemispherical Surface
The area \( A \) of a hemispherical surface is given by:
\[ A = 2\pi R^2 \]
Substituting \( R = 1.2 \, \text{m} \):
\[ A = 2\pi (1.2)^2 = 2\pi (1.44) = 2.88\pi \, \text{m}^2 \]

### Step 4: Determine the Angle \( \theta \)
For a hemispherical surface, the electric field is perpendicular to the flat circular base of the hemisphere. Thus, for the curved surface, the angle \( \theta = 0^\circ \) (the electric field is parallel to the area vector of the curved surface).

### Step 5: Calculate the Electric Flux
Since \( \cos(0^\circ) = 1 \), the electric flux through the hemispherical surface is:
\[ \Phi_E = E \cdot A \cdot \cos(0^\circ) = E \cdot A \]
Substituting the values:
\[ \Phi_E = 500 \, \text{N/C} \cdot 2.88\pi \, \text{m}^2 \]
\[ \Phi_E = 1440\pi \, \text{N m}^2/\text{C} \]

### Step 6: Express in the Form \( N\pi \)
From the calculation, we have:
\[ \Phi_E = 1440\pi \]
Thus, comparing with the expression \( N\pi \), we find that:
\[ N = 1440 \]

### Final Answer
The value of \( N \) is \( \boxed{1440} \).

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