Step 1: Apply Gauss's Law. Gauss's Law states that the net electric flux through any closed surface is proportional to the net electric charge enclosed by that surface (\( \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \)).
Step 2: Consider a Gaussian surface inside the hollow conductor. For a hollow spherical conductor, any charge placed on it will reside entirely on its outer surface due to electrostatic repulsion. If we draw a spherical Gaussian surface at any radius \(r\) inside the hollow conductor, this surface encloses no charge (\(Q_{enc} = 0\)).
Step 3: Conclude the value of the electric field. Since \(Q_{enc} = 0\), Gauss's Law gives: \[ \oint \vec{E} \cdot d\vec{A} = 0 \] Due to the spherical symmetry, the electric field \(E\) must be constant in magnitude on the Gaussian surface and directed radially. Therefore, the integral becomes \( E \cdot (4\pi r^2) = 0 \). Since the area \(4\pi r^2\) is not zero, the electric field \(E\) must be zero everywhere inside the conductor.
If \(f(t)\) is the inverse Laplace transform of \( F(s) = \frac{s+1+s^{-2}}{s^2-1} \), then \(f(t)\) is
Match LIST-I with LIST-II
LIST-I (Differential Equation)
(A) \(\frac{dy}{dx} = 2x(y-x^2+1)\)
(B) \(x\frac{dy}{dx} + 2(x^2+1)y=6\)
(C) \((x^2+1)\frac{dy}{dx} + 2xy = x \sin x\)
(D) \(x^3\frac{dy}{dx} + 2xy = 2x^2e^{x^2}\)
LIST-II (Integrating Factor)
(I) \(x^2\)
(II) \(e^{-x^2}\)
(III) \(x^2e^x\)
(IV) \(1+x^2\)
Choose the correct answer from the options given below: