Step 1: Apply Gauss's Law. Gauss's Law relates electric flux to enclosed charge: \( \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \).
Step 2: Consider a Gaussian surface within the hollow conductor. In a hollow spherical conductor, charge resides on the outer surface. A spherical Gaussian surface inside encloses no charge (\(Q_{enc} = 0\)).
Step 3: Determine the electric field. With \(Q_{enc} = 0\), Gauss's Law yields \[ \oint \vec{E} \cdot d\vec{A} = 0 \]. Due to symmetry, \(E\) is constant and radial. Thus, \( E \cdot (4\pi r^2) = 0 \). Since the area isn't zero, the electric field \(E\) is zero inside the conductor.
A point charge \(q = 1\,\mu\text{C}\) is located at a distance \(2\,\text{cm}\) from one end of a thin insulating wire of length \(10\,\text{cm}\) having a charge \(Q = 24\,\mu\text{C}\), distributed uniformly along its length, as shown in the figure. Force between \(q\) and wire is ________ N. 