Let the electrostatic field E at distance r f | vedclass

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(B) Statement $(I)$ is invalid because Gauss's law $(\phi = q_{	ext{enclosed}} / \varepsilon_0)$ is only valid for inverse-square fields. If $E = \fr

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Only statement $II$ is valid

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(B) Statement $(I)$ is invalid because Gauss's law $(\phi = q_{	ext{enclosed}} / \varepsilon_0)$ is only valid for inverse-square fields. If $E = \frac{kq}{r^3} \hat{r}$, the flux through a spherical surface of radius $r$ is:$\phi = \oint E \cdot dA = \int \frac{kq}{r^3} dA = \frac{kq}{r^3} (4\pi r^2) = \frac{4\pi kq}{r} 
eq \frac{q}{\varepsilon_0}$.Statement $(II)$ is valid. In an inverse-square field, the electric field inside a uniformly charged shell is zero due to symmetry. However, for an inverse-cubic field, the superposition of fields from different parts of the shell does not cancel out at interior points. Thus, a charge placed inside will experience a non-zero net force.Therefore, only statement $(II)$ is valid.

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