A long cylindrical shell carries positive surface charge $$\sigma$$ in the upper half and negative surface charge $$-\sigma$$ in the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale)

We start from the very basic rule for electrostatic field lines: they originate on positive charge and they terminate on negative charge. Hence, wherever the surface charge density is $$+\sigma$$ the field lines must leave the surface, and wherever it is $$-\sigma$$ they must enter the surface.

Another fundamental fact, valid for any conductor in electro-static equilibrium, is the relation between the surface charge density $$\sigma$$ and the electric field just outside the conductor:

$$\mathbf E_{\text{out}}=\dfrac{\sigma}{\varepsilon_0}\,\hat n,$$

where $$\hat n$$ is the unit vector normal to the surface directed outward from the metal. Therefore

• for $$\sigma>0$$ (upper half of the cylindrical shell) the field just outside the metal points radially outward, normal to the surface,
• for $$\sigma<0$$ (lower half of the cylindrical shell) we have $$-\sigma<0$$, so the field just outside points radially inward, again normal to the surface.

Because the shell is conducting, the electric field inside the metal itself is zero; hence no field lines can exist inside the cylindrical material, nor in the hollow region surrounded by it.

Putting these two pieces together we get the complete qualitative picture:

• Lines emerge perpendicularly from every point on the outer surface of the upper (positively charged) half.
• They terminate perpendicularly on the outer surface of the lower (negatively charged) half.
• No lines cross the metal, and no lines appear inside the cavity.

Among the four schematic figures supplied in the question, only Figure (2) shows lines that leave the upper semicylinder, curve through the external space, and enter the lower semicylinder, while the interior of the shell remains field-free. The other three options either show lines entering/leaving the wrong halves or show field lines existing inside the conductor, both of which violate the rules stated above.

Hence, the correct answer is Option B (Figure 2).