By Gauss's law, the electric field at a perpendicular distance $r$ from an infinitely long straight wire with uniform linear charge density $\lambda$ is given by:

For a charged conducting spherical shell of radius $R$ carrying charge $q$:

• Inside the shell ($r \lt R$): The charge enclosed by any Gaussian surface inside the shell is zero, so $E = 0$.
• At the outer surface ($r = R$): The electric field abruptly reaches its maximum value: $E = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2}$.
• Outside the shell ($r \gt R$): The electric field decreases inversely as the square of distance: $E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \propto \frac{1}{r^2}$.

Graph 2 correctly shows $E = 0$ for $0 \le r \lt R$, a discontinuous vertical jump at $r = R$, followed by an inverse-square decay curve for $r \gt R$.

For any point on or outside a spherical charged conductor at distance $r$ from the centre:

Taking the ratio of $V$ and $E$:

Given:

• $E = 10\,\text{V/m}$
• $r = 20\,\text{cm} = 0.20\,\text{m}$

Copper is a metal (conductor) with an infinite dielectric constant ($K \to \infty$).

The capacitance of a parallel plate capacitor with a slab of thickness $b$ and dielectric constant $K$ is:

For a metallic conductor like copper, $K \to \infty$, which gives $\frac{b}{K} \to 0$:

The effective separation between the capacitor plates is reduced from $d$ to $(d - b)$. Hence, option 2 is the correct answer.