In 1974, Sir Augusts de coulomb studied the relationship between localized charges. He carried out experiments using torsion balance. On the basis of his experimental results, he proposed a law known as Coulomb’s law.

Consider two point charges q₁ and q₂ and let "r" be the separation between them. According to Coulomb’s law:

F µ q₁ q₂ -----------(i)

Where K is the constant of proportionality and its value depends upon the medium between the two charges. In S.I system, its value is equal to:

K = 8.98 x 10⁹ N-m² / c²

Where Î _o is the permitivity of free space and its value is:

Î _o = 8.85 x 10^-12 col² / Nm²

Where Î _r is the relative permitivity which is different for different dielectric.

It is the measure of the strength of the electric field. Electric intensity is a vector quantity and its direction is same as that of force. If the charge is positive then electric intensity is directed from the charge and if the charge is negative, then it is directed towards the charge. The SI unit of electric intensity is Newton per Coulomb or volt per meter.

Mathematically electric intensity is given by:

Where:

E = Electric Intensity

F = Force

q° = small magnitude

Consider a charged body q that has an electric field all around it. We want to find out electric intensity E at point P. For this purpose, we placed a point charge q₁ at that point.

The flux at a surface is determined by the product of flux density i.e. electric field and the projection of its area perpendicular to the field or by the product of area and component of field normal to the area.

Mathematically,

D j _e = D A (Ecosq )

Thus flux is a scalar product of electric intensity and vector Area. Although area is a scalar quantity, D A is taken as a vector quantity of magnitude equal to area and the direction normal to it.

Electric flux will be maximum when electric lines of forces and normal area are in the same direction.

Mathematically,

D j _e = D A (Ecosq )

But q = 0° ,

D j _e = D A (Ecos0)

Since cos 0 = 1, therefore,

D j _e = ED A

Þ D j _eMAX = ED A

Electric Flux will be minimum when electric lines of forces and normal area are perpendicular to each other.

D j _e = D A (Ecosq )

But q = 90° ,

D j _e = D A (Ecos90)

Since cos 90 = 0, therefore,

D j _e = 0

Þ D j _eMIN =0

Consider an isolated point charge +q. The lines of forces from q will spread uniformly in space around it cutting the surface of an imaginary sphere. Now we want to find flux due to point charge. For this purpose, we divide the whole sphere into small patches. Each patch is denoted by D A.

We known that,

D j _e = D A (Ecosq )

But q = 0° ,

D j _e = D A (Ecos0)

Since cos 0 = 1, therefore,

D j _e = ED A

For the whole sphere,

S j _e = S E. D A

Þ j _e = E S D A-----------(I)

Since,

D A = 4p r²

According to Guass’s law,

Where Î o is the permitivity of free space and its value is 8.85 x 10^-12 col²/Nm².

Consider a closed surface ‘s’ in which q₁, q₂, ……, q_n point charges are present. Now we want to find the total electric flux due to all these point charges.

As we know that electric flux due to point charge is q/Î _o, therefore,

Electric Flux due to q₁ = j ₁ = q₁/Î _o

Electric Flux due to q₃ = j ₂ = q₂/Î _o

Electric Flux due to q₃ = j ₃ = q₃/Î _o

Electric Flux due to q_n = j _n = q_n/Î _o

Since electric flux is a scalar quantity, therefore, for total electric flux we will add those individual electric fluxes.

j _e = j ₁ + j ₂ + j ₃ + ….. + j _n

Þ j e = q₁/Î _o + q₂/Î _o + q₃/Î _o + q_n/Î _o

Þ j e = 1/ Î _o (q₁ + q₂ + q₃ + q_n)

Þ j _e = 1/Î _o (Total Charge)

Þ j _e = 1/Î _o (Q)

Guass’s law can be used to calculate the electric field only in those cases of charge distribution which are so symmetrical that by proper choice of Guassian surface the flux on it may possibly be evaluated. With the help of guass’s law we can measure the intensity in following cases:

1. Electric Intensity due to charge sheet
2. Electric Intensity due to two charge sheets

Consider a charge sheet in which unit positive charges are uniformly distributed

As we know that charge density is the charge stored per unit area and is denoted by:

Þ Q = s

Now consider a cylindrical shell, which is placed inside the charge sheet. It has three surfaces:

• Upper Surface
• Curve Surface
• Bottom

j ₁ = E. D A

j ₁ = ED A cos q

But q = 0° and cos 0° = 1, therefore,

j ₁ = E D A cos 0°

Þ j ₁ = E D A

j ₂ = E. D A

j ₂ = ED A cos q

Since the angle between the field vector and area vector of all elements of curved surface is 90° , therefore,

j ₂ = E D A cos 90°

But cos 90° = 0,

j ₂ = E D A (0)

Þ j ₂ = 0

j ₃ = E. D A

j ₃ = ED A cos q

But q = 0° and cos 0° = 1, therefore,

j ₃ = E D A cos 0°

Þ j ₃ = E D A

Total Flux is given by:

j _e = j ₁ + j 2 + j ₃

Þ j _e = ED A + 0 + ED A

• j _e = 2ED A

For the whole charge sheet:

• j _e = 2E S D A

• j e = 2EA ------------- (II)

Capacitor is a device which is use to store charge. A simple capacitor consists of two parallel metallic plates. A plate is connected to the positive terminal of the battery and another plate is connected to the negative terminal of the battery. As shown in the following figure.

If V is the voltage provided to the capacitor and Q is the amount of charge stored in the capacitor, then it is observed that if more is the voltage, then more will be the charge stored in the capacitor.

Mathematically,

Q µ V

Þ Q = CV

Where C is the capacitance of the capacitor which may be defined as,

The unit of Capacitance is Farad. It may be defined as,

The capacitance of a capacitor depends upon following factors:

1. Cross-sectional area of plate
2. Separation between Plates
3. Dielectric

On the basis of dielectric capacitors are classified into different types. For example Electrolytic Capacitor, Paper Capacitor, Meca Capacitor ,Oil Capacitor.

There are two main types of capacitors:

1. Fixed Capacitor
2. Variable Capacitor

Those capacitor whose capacitance is not fixed are known as Variable capacitor. For example Gang Capacitor.

Consider a parallel plate capacitor in which d is the distance between the plates, the charge stored is denoted by Q where as potential difference between the plates is V. Consider the following figure.

As we known that electric intensity due to two charge sheets is:

According to the definition of electrical potential:

D V = E.D r

Substituting the value of E from eq(I) in above equation:

Capacitors of some fixed values are used in a circuit. The capacitance of a desired value can however be obtained by suitable combination of capacitor. Capacitors can be combined in parallel, series or both.

Consider three capacitors having capacitances C₁, C₂, C₃ connected in a series. These capacitors can be replace by an equivalent capacitor having capacitance C_e. When a cell is connected across the ends of system then a charge Q is transferred across the plates of capacitors. A charge upon one plate always attracts upon the other plate with a charge equal in magnitude and opposite in sign.

Let V be the potential difference across the combination. The potential difference across the individual capacitor is V_ab, V_bc, V_cd.

As we know that in case of series combination the charge across the individual capacitor remains constant, where as potential difference varies such that the potential difference V is the sum of potential difference applied across individual capacitor.

V_ad = V_ab + V_bc + V_cd ----------- (I)

The reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances.

Consider three capacitors having C₁, C₂, C₃ capacitances respectively are connected in parallel. We can replace them by an equivalent capacitor having capacitance C_e. A charge q given to a point divide it self reside on the plates of individual capacitor as Q₁, Q₂, Q₃ respectively.

As we known that in case of parallel combination the potential difference across each capacitor is that of the source where as the charge across each capacitor varies, therefore the total charge Q is given by:

Q = Q₁+ Q₂ + Q₃ ---------- (I)

As we known that,

Q = CV

Therefore,

Q = C_e V_ab

Þ Q₁ = C₁ V_ab

Þ Q₂ = C₂ V_ab

Þ Q₃ = C₃ V_ab

Substituting the values of Q₁, Q₂, Q₃ in eq (I)

C_eV_ab = C₁V_ab + C₂V_ab + C₃V_ab

C_eV_ab = V_ab( C₁ + C₂ + C₃)

C_e = C₁ + C₂ + C₃

The equivalent capacitance is equal to sum of the individual capacitances.

In order to bring a unit positive charge from one point to another point, we have to do some work. This work is stored in the form of potential energy. This potential energy per unit charge is known as Electric Potential.

Electric Potential is denoted by V. It is defined as the potential energy per unit charge. Therefore mathematically,

We know that:

D u = F.D r

But, E = F/q Þ F = Eq, therefore:

D u = EqD r

Substituting the value in eq(I)

Þ D V = E.D r

Where,

D V = Electric Potential

E = Electric Intensity

D r = Displacement

From above expression we know that:

It is a scalar quantity and its unit is Volt.

Electric Potential is the amount of work done in order to move a unit positive charge from one point to another against the direction of electric.

Consider a unit positive charge placed in a uniform electric field. We have to displace it from point O to point N. For this purpose we have to do some work. This work is known as electric potential. In order to determine electric potential from point O to point N we divide the whole distance into small equal patches because in a long distance intensity does not remain constant. Each patch is denoted by D r.