Please help me solve this promobles

The electric field intensity $E$ at a distance $r$ from a point charge $Q$ is given by:

$$E = \frac{kQ}{r^2}$$

where $k$ is Coulomb's constant, $k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$.

Given $Q = 5 \times 10^{-6} \, C$ and $r = 0.5 \, m$.

$$E = \frac{(9 \times 10^9 \, \text{Nm}^2/\text{C}^2) \times (5 \times 10^{-6} \, C)}{(0.5 \, m)^2}$$

$$E = \frac{45 \times 10^3 \, \text{Nm}^2/\text{C}}{0.25 \, m^2}$$ $$E = 180 \times 10^3 \, \text{N/C}$$ $$E = 1.8 \times 10^5 \, \text{N/C}$$

The electric field intensity is $\boxed{\text{}1.8 \times 10^5 \, \text{N/C}\text{}}$

Question 22:

a) Define electric potential.
Electric potential at a point in an electric field is defined as the amount of work done per unit positive test charge in bringing the charge from infinity to that point without acceleration. It is a scalar quantity.
Its SI unit is joules per coulomb ($\text{J/C}$) or volts ($\text{V}$).

b) Calculate the electric potential at a point $0.5 \, m$ away from a point charge of $5 \times 10^{-6} \, C$.

The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by:

$$V = \frac{kQ}{r}$$

where $k$ is Coulomb's constant, $k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$.

Given $Q = 5 \times 10^{-6} \, C$ and $r = 0.5 \, m$.

$$V = \frac{(9 \times 10^9 \, \text{Nm}^2/\text{C}^2) \times (5 \times 10^{-6} \, C)}{0.5 \, m}$$

$$V = \frac{45 \times 10^3 \, \text{Nm/C}}{0.5}$$ $$V = 90 \times 10^3 \, \text{V}$$ $$V = 9.0 \times 10^4 \, \text{V}$$

The electric potential is $\boxed{\text{}9.0 \times 10^4 \, \text{V}\text{}}$

Question 23:

a) Define electric flux.
Electric flux is a measure of the number of electric field lines passing through a given surface. It quantifies the flow of the electric field through an area.
It is given by the dot product of the electric field and the area vector: $\Phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta$.
Its SI unit is newton-meter squared per coulomb ($\text{Nm}^2/\text{C}$) or volt-meter ($\text{Vm}$).

b) State Gauss's Law.
Gauss's Law states that the total electric flux through any closed surface (a Gaussian surface) is directly proportional to the total electric charge enclosed within that surface.
Mathematically, it is expressed as:

$$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$

where $\Phi_E$ is the total electric flux, $\vec{E}$ is the electric field, $d\vec{A}$ is an infinitesimal area vector on the closed surface, $Q_{enc}$ is the total charge enclosed by the surface, and $\epsilon_0$ is the permittivity of free space.

Question 24:

a) Define electric current.
Electric current is defined as the rate of flow of electric charge through a conductor. It measures how much charge passes through a given point in a circuit per unit time.
Its SI unit is amperes ($\text{A}$), where $1 \, \text{A} = 1 \, \text{C/s}$.

b) Calculate the number of electrons passing through a point in a conductor in $1 \, minute$ if the current is $2 \, A$.

The relationship between current ($I$), charge ($Q$), and time ($t$) is $Q = I \times t$.
Given $I = 2 \, A$ and $t = 1 \, \text{minute} = 60 \, s$.

$$Q = 2 \, A \times 60 \, s$$ $$Q = 120 \, C$$

The charge of a single electron is $e = 1.6 \times 10^{-19} \, C$.
The number of electrons $N$ is the total charge divided by the charge of one electron:

$$N = \frac{Q}{e}$$ $$N = \frac{120 \, C}{1.6 \times 10^{-19} \, C/\text{electron}}$$

$$N = 75 \times 10^{19} \, \text{electrons}$$ $$N = 7.5 \times 10^{20} \, \text{electrons}$$

The number of electrons passing through the point is $\boxed{\text{}7.5 \times 10^{20} \, \text{electrons}\text{}}$