# Electric Charge and Electric Field – Electric Flux, Coulomb’s Law, Sample Problems

The word “*electricity*” comes from the Greek word “Elektron” which means “*amber*“. The magnetic and electric forces present in materials, atoms, and molecules affect their properties. The term “electric charge” refers to just two types of entities. An experiment revealed two forms of electrification: first, the like charges that repel one another, and other is unlike charges that attract one another. The polarity of charge is the distinguishing element between these two sorts of charges.

An experiment on electric charges generated by frictional electricity revealed that conductors help in the passage of electric charge whereas insulators do not. Metals, the Earth, and human bodies are all conductors, whereas porcelain, nylon, and wood are all insulators, providing substantial resistance to the flow of electricity through them.

### Electric Field

An electric charge creates an electric field, which is a region of space around an electrically charged particle or object where the charge feels forced. The electric field exists everywhere in space and can be studied by introducing another charge into it.

If the charges are far enough apart, the electric field can be approximated as 0 for practical purposes. Electric fields are a** **vector** **quantity represented by arrows pointing toward or away from charges. The lines must point radially outward, away from a positive charge, or radially inward, toward a negative charge.

The electric field’s magnitude is calculated using the formula:

**E = F/q**

- where E is the electric field’s strength,
- F is the electric force,
- q is the test charge.

### Properties of Electric Field Lines

- Lines of the electric field are continuous curves.
- They begin with a positively charged body and end with a negatively-charged one.
- The direction of electric field intensity at any point is determined by being tangent to the electric field line.
- There are no two electric field lines that cross each other.
- Electric field lines are always parallel to the conductor’s surface.

### Coulomb’s Law

The force that exists between two point charges is described by Coulomb’s Law. In physics, the phrase point charge refers to the fact that linearly charged objects have a small size in contrast to the distance between them. As a result, we treat them as point charges since calculating the attraction/repulsion force between them is straightforward.

The statement, in general, includes two charges, q

_{1}and q_{2}. The attraction/repulsion force between the charges is denoted by the letter ‘F,’ while the distance between them is denoted by the letter ‘r.’ Then Coulomb’s law is expressed mathematically as-

- F is proportional to the product of the magnitudes of the charges in contact, i.e. F ∝ q
_{1}q_{2}.- F is inversely proportional to the square of the distance between the two charges in contact, i.e. F ∝ 1/ r
^{2}.

Let’s put the two together as follows:

**F ∝ q _{1}q_{2} / r^{2}**

Now, if we remove proportionality a constant k is introduced;

**F = k q _{1}q_{2}/ r^{2}**

where k is the proportionality constant and equals 1/40, and 0 is the epsilon not, which denotes the vacuum permittivity. k has been calculated to be 9 × 10^{9} Nm^{2}/ C^{2}.

According to Coulomb, like charges repel one other whereas unlike charges attract each other. This indicates that charges of the same sign repel one another, whereas charges of opposite signs attract one another.

### Electric Flux

The electric flux is the total number of electric field lines moving through a particular area in a given amount of time. However, unlike in the instance of liquid flow, there is no flow of a physically observable amount.

To begin with, the definition of electric flux through an area element S is as follows:

**Δθ= E.ΔS= E ΔS cosθ**

The amount of field lines cutting the area element determines this. The angle, in this case, is the angle formed by E and S. In a closed surface, when the convention has previously been established, is the angle formed by E and the area element’s outward normal. To determine the overall flux through a given surface, divide it into small area elements, compute the flux at each element, then add them together. As a result, the total flow through a surface S equals E.S. Because the electric field E is assumed to be constant over the small area element, the approximation symbol is employed.

**Electric Dipole:** It consists of a pair of equal or opposite charges A and -B separated by 2x. The dipole moment vector has a magnitude of 2Ax and points from -B to A in the direction of the dipole axis.

### Sample Questions

**Problem 1: A charge of 2 C is placed in the center of an 8 cm ^{3} cube. What is the magnitude of the electric flux traveling through one of the faces?**

**Solution:**

Because the given cube’s volume is 8 cm3, its side length is 2 cm.

Electric flux over a closed surface is contained charge within the closed surface divided by the medium’s permittivity, according to Gauss’s theorem.

If the medium within the cube is air or vacuum, the total electric flux over the closed surface is

= Q / Eo

= 2 / ( 8.854 10

^{-12})= 2.259 10

^{11}N C^{-1}m^{2}.Electric flux passing through one face = ( 2.259 × 10

^{11}) / 6 = 3.765 × 10^{10}N C^{-1}m^{2}.

**Problem 2: Why do two electric field lines never intersect?**

**Solution:**

There will be two tangents and consequently two directions of net electric field at the point where the two lines join, which is not possible. As a result, two electric field lines do not cross.

**Problem 3: A force of 8 N is experienced when two point charges separated by 1 m have equal charges. What force will they feel if they are both submerged in water at the same time? (Assuming K water = 80)**

**Solution:**

Force acting between two point charges

F air=q1q2/4πEoxr2

F water=q1q2/4πEoKxr2

Therefore ,

F air/F water=K

8/F water=80

F water=8/80

=1/10N

**Problem 4: How does the electric field inside a dielectric decrease when it is placed in an external electric field?**

**Solution:**

When a dielectric is exposed to an electric field (E vector), charge is induced in it, producing an electric field in the opposite direction of E. As a result, the net electric field is lowered.

E

_{net }= E vector −Epwhere Ep denotes the field produced by dielectric polarisation.

**Problem 5: Consider a system of two charges of magnitude 3 × 10-7 C and 4 × 10-7 C which are acted upon by a force of 0.1 N. What is the distance between the two charges?**

**Solution:**

Given that,

The first charge, q

_{1}is 3 × 10^{-7}C.The second charge, q

_{2 }is 4 × 10^{-7}C.The force acted upon them, F is 0.1 N.

The formula to calculate electrostatic force between the charges is:

F = k q

_{1}q_{2 }/ r^{2}Substitute the given values in the above expression as,

0.1 N = (9 × 10

^{9}Nm^{2}/ C^{2})(3 × 10^{-7}C)(4 × 10^{-7}C) / (r)^{2}r = 0.103 m

Hence, the distance between the two charges, r is 0.103m.

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