For two large parallel sheets carrying surface charge densities $+\sigma$ and $-\sigma$:

• Outside the sheets (regions at $P_1$ and $P_3$), the electric fields produced by the two sheets are equal in magnitude ($E = \frac{\sigma}{2\varepsilon_0}$) and opposite in direction. They cancel completely, yielding $\vec{E}_1 = 0$ and $\vec{E}_3 = 0$.
• Between the sheets (at $P_2$), the electric fields produced by both sheets point in the same direction (from positive sheet towards negative sheet) and add up:
  $$\vec{E}_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \neq 0$$
  Since electrostatic force on charge $q$ is $\vec{F} = q \vec{E}$:
  $$\vec{F}_1 = 0,\quad \vec{F}_2 \neq 0,\quad \vec{F}_3 = 0$$

Using Gauss's law with a cylindrical Gaussian surface of cross-sectional area $A$ piercing perpendicularly through an infinite thin plane sheet with uniform surface charge density $\sigma$:

The field is uniform, independent of distance from the sheet, and directed perpendicularly outward (for positive $\sigma$).

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 $r \gt R$.