Numerical Problems Based on Applications of Gauss’s Law for Class 12 Physics

Given Electric Field $E = 9 \times 10^4\text{ NC}^{-1}$ and distance $r = 4\text{ cm} = 0.04\text{ m}$.

The electric field due to an infinite line charge is given by $E = \frac{\lambda}{2\pi\varepsilon_0 r}$.

We know that $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ Nm}^2\text{C}^{-2}$, so $\frac{1}{2\pi\varepsilon_0} = 18 \times 10^9\text{ Nm}^2\text{C}^{-2}$.

For a cylinder of large length, it behaves like an infinite line charge outside its surface.

Given linear charge density $\lambda = 2 \times 10^{-8}\text{ Cm}^{-1}$ and distance $r = 0.2\text{ m}$.

First, convert the linear charge density to standard SI units ($\text{C/m}$):

Given distance $r = 4\text{ m}$ (the height above the earth). Applying the formula for the electric field of a line charge:

(Note: If the original answer was given as $4.5 \times 10^3\text{ Vm}^{-1}$, it implies the charge was $1\text{ }\mu\text{C}$ per meter rather than per cm. Mathematically, for per cm, it evaluates to $10^5$).

The electric field between two parallel, oppositely charged plates is uniform and is given by $E = \frac{\sigma}{\varepsilon_0}$.

Surface charge density is $\sigma = \frac{q}{A}$. Substituting this into the field equation gives:

Given $E = 100\text{ NC}^{-1}$, $A = 1\text{ m}^2$, and permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$:

The necessary centripetal force for the electron’s revolution is provided by the electrostatic force of attraction between the electron and the line charge.

We know the electric field of a line charge is $E = \frac{\lambda}{2\pi\varepsilon_0 r}$. Substitute this into the force equation:

Multiplying both sides by $r$ shows that $mv^2$ is indeed independent of the radius $r$:

Kinetic Energy is $K.E. = \frac{1}{2}mv^2$:

Now, plug in the values ($k = 9 \times 10^9$, $e = 1.6 \times 10^{-19}\text{ C}$, $\lambda = 2 \times 10^{-8}\text{ C/m}$):