A hemispherical shell is uniformly charge positively. The electric field at point on a diameter away from the center is directed

Options

A. perpendicular to the diameter
B. parallel to the diameter
C. at an angle tilted towards the diameter
D. at an angle tilted away from the diameter.

Explanation

To find the direction of the electric field at a point on the diameter of a uniformly positively charged hemispherical shell, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a hemispherical shell that is uniformly charged positively.
- We need to find the electric field at a point on the diameter of the shell, away from the center.

2. **Identify the Point of Interest**:
- Let's denote the center of the hemispherical shell as point O and the point on the diameter where we want to find the electric field as point A.

3. **Consider the Symmetry**:
- Due to the symmetry of the hemispherical shell, we can consider an elemental ring of charge on the surface of the shell.
- Each elemental charge on this ring will contribute to the electric field at point A.

4. **Electric Field Contribution from the Ring**:
- The electric field due to a ring of charge at a point along its axis (which is also the diameter in this case) is directed away from the ring if the charge is positive.
- The electric field \( E \) at a distance \( x \) from the center of the ring can be expressed as:
\[ E = \frac{k \cdot dQ \cdot x}{(x^2 + R^2)^{3/2}} \]
where \( dQ \) is the charge on the ring, \( R \) is the radius of the ring, and \( k \) is Coulomb’s constant.

5. **Components of the Electric Field**:
- Each elemental charge \( dQ \) on the ring produces an electric field component directed away from the ring.
- The components of the electric field that are parallel to the diameter will cancel out due to symmetry (as there are equal charges on either side of the diameter).
- The components of the electric field that are perpendicular to the diameter will add up.

6. **Resultant Electric Field Direction**:
- Since the parallel components cancel out, the net electric field at point A will be directed perpendicular to the diameter of the hemispherical shell.
- Given that the shell is positively charged, the direction of the electric field will be away from the shell.

7. **Conclusion**:
- Therefore, the electric field at point A on the diameter away from the center is directed perpendicular to the diameter of the shell.

### Final Answer:
The electric field at the point on the diameter away from the center is directed **perpendicular to the diameter**.

---