A cone of base radius R and height h is located in a uniform electric field `vecE` parallel ot its base. The electric flux entering the cone is :

Options

A. `1/2 Eh R`
B. `2EHR`
C. `4EhR`
D. `E h r`

Answer

To find the electric flux entering a cone placed in a uniform electric field parallel to its base, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
Electric flux (Φ) is defined as the dot product of the electric field (E) and the area vector (A) through which the field lines pass. Mathematically, it is given by:
\[ \Phi = \vec{E} \cdot \vec{A} = E A \cos \theta \]
where θ is the angle between the electric field vector and the area vector.

2. **Identify the Geometry of the Cone**:
- The cone has a base radius \( R \) and height \( h \).
- The electric field \( \vec{E} \) is parallel to the base of the cone.

3. **Determine the Area Vector**:
- The area vector for the base of the cone is directed perpendicular to the base. Therefore, it is at a 90-degree angle to the electric field.
- For the base of the cone, \( \theta = 90^\circ \).

4. **Calculate Flux Through the Base**:
- Since \( \cos(90^\circ) = 0 \), the electric flux through the base of the cone is:
\[ \Phi_{\text{base}} = E A \cos(90^\circ) = 0 \]
Thus, the base does not contribute to the electric flux.

5. **Calculate the Area of the Curved Surface**:
- The curved surface of the cone can be approximated as two right-angled triangles. The area \( A \) of the curved surface can be calculated as:
\[ A = 2 \times \left(\frac{1}{2} \times R \times h\right) = R \times h \]

6. **Determine the Angle for the Curved Surface**:
- For the curved surface, the area vector is parallel to the electric field, so \( \theta = 0^\circ \).
- Therefore, \( \cos(0^\circ) = 1 \).

7. **Calculate the Electric Flux Through the Curved Surface**:
- The electric flux through the curved surface is given by:
\[ \Phi_{\text{curved}} = E A \cos(0^\circ) = E (R \times h) \times 1 = E R h \]

8. **Final Result**:
- The total electric flux entering the cone is:
\[ \Phi = E R h \]

### Conclusion:
The electric flux entering the cone is given by:
\[ \Phi = E R h \]