Electrostatics Notes

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CHAPTER – 12                                                                                                    ELECTROSTATICS

Qs. State and Explain Coulomb’s Law.

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

Statement
The electric force of interaction between two point charges is directly proportional to product of their charges and inversely proportional to the square of the distance from their centers.

Mathematical Expression
Consider two points charges q1 and q2 and let “1″ be the separation between them. According to Coulomb’s Law.
F ∞ q1 q2 ——– (i)
And, according to second part
F ∞ 1/r² ——— (ii)
Combining eq (i) and eq (ii), we get
F ∞ q1 q2 / r²
F = K q1 q2 / r²
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 10(9) N-m² / c².
Coulomb’s Law can be expressed in terms of permitivity as follows.
F = 1 / 4 π Є . q1 q2 / r²
Where Єo is the permitivity of free space and its value is
Єo = 8.85 x 10(-12) col² / Nm²
if some other medium is used instead of air then
F = 1 / 4 π Єo Єr . q1 q2 / r²
Where Єr is the relative permitivity which is different for different dielectric.


Qs. Define electric field and electric intensity. Find the formula for the electric intensity due to a point charge.

ELECTRIC INTENSITY

DEFINITION
Electric intensity is the force experienced by a unit positive charge due to the presence of a charge body.

EXPLANATION
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.

Mathematical Form
Mathematically electric intensity is given by
E = F / q°
Where E = Electric Intensity
F = Force
q° = small magnitude

ELECTRIC INTENSITY DUE TO A POINT CHARGE

CONSIDERATION
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.

Qs. Define Electric flux. Find the formula for the electric flux due to point charge in a closed sphere.

ELECTRIC FLUX

DEFINITION
The total number of imaginary lines drawn in such a way that the attraction of tangent at any point is same as that of electric field crossing a surface normally is called electric flux or flux on the surface.
OR
The total number of lines of forces crossing a surface normally is called flux on that surface.

Mathematical Form
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.

ELECTRIC FLUX DUE TO A POINT CHARGE IN A CLOSED SPHERE
Consider an isolated point charge +q. Th 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 ΔA.

Qs. State and prove Guass’s Law

The total electric flux diverging out from a closed surface is equal to the product of the sum of all charges present in that closed surface and 1/ Єo.

Find the Formula for Electric Intensity.
1. Due to Charge Sheet
2. Due to Two Charge Sheets
OR
What are the application of Guass’s Law.

Introduction
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.

1. Electric Intensity Due to Charge Sheet.
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
σ = Q / A
=> Q = σ A
According to Guass’s Law
σ = Q / Є
=> φe = σ A / Є ——– (I)
Now consider a cylindrical shell, which is placed inside the charge sheet. It has three surfaces.
Upper Surface
Curve Surface
Bottom

Electric Intensity Due to Upper Surface
φ1 = E ΔA
φ1 = E ΔA cos θ
But θ = 0° and cos 0° = 1, therefore,
φ1 = E ΔA cos 0°
= φ1 = E ΔA

Electric Intensity Due to Curved Surface
φ2 = E ΔA
φ2 = E ΔA cos θ
Since the angle between the field vector and area vector of all elements of curved surface is 90°, therefore,
φ2 = E ΔA cos 90°
But cos 90° = 0,
φ2 = E ΔA (0)
φ2 = 0

Electric Intensity Due to Lower Surface
φ3 = E ΔA
φ3 = E ΔA cos θ
But 0 = 0° and cos 0°, = 1 therefore,
φ3 = E ΔA cos 0°
=> φ3 = E ΔA

Total Flux
Total Flux is given by
φe = φ1 + φ2 + φ3
=> φe = E ΔA + 0 + E ΔA
=> φe = 2E ΔA
For the whole charge sheet:
=> φe = 2E Σ ΔA
=> φe = 2 EA ———– (II)
Comparing equation (I) and (II)
=> σ A / Єo = 2AE
=> E = σ / 2 Єo
OR
E = σ / 2 Єo

2. Electric Intensity Due to Two Charge Sheets.
As we know that electric intensity due to a charge sheet is
E = σ / 2 Єo
For two sheets E will be,
=> E = σ / 2 Єo + σ / 2 Єo
=> E = σ + σ / 2 Єo
=> E = 2 σ / 2 Єo
=> E = σ / Єo
=> E = σ / Єo i

Electrostatics MCQ with answers

Qs. Define the term Capacitor and Capacitance of a Capacitor

CAPACITOR
Capacitor is a device which is use to storage 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.

CAPACITANCE

Definition
The capacity of a capacitor to store the charge is known as Capacitance.

Mathematical Explanation
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 ∞ V
=> Q = CV
Where C is the capacitance of the capacitor which may be defined as
The ratio of the charge on one of the plate (conductor) to the potential difference between them.

Unit of Capacitance
The unit of Capacitance is Farad. It may be defined as,
If one coulomb charge is stored due to 1 volt potential, then capacitance will be 1 Farad.

Factors on which Capacitance Depends.
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 Electrolyte Capacitor, Paper Capacitor, Meca Capacitor, Oil Capacitor.

Types of Capacitor
There are two main types of Capacitors.
1. Fixed Capacitor
2. Variable Capacitor

1. Fixed Capacitor
Those capacitor whose capacitance is constant are known as Fixed Capacitor. For example Paper Capacitor and Meca Capacitor.

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

Find the Formula for the capacitance of Parallel plate capacitor.

PARALLEL PLATE CAPACITOR

Definition
A parallel plate capacitor is a device used to store the charge. It consist 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.
These plates are separated by a very small distance compared to the dimension of the plates.

CAPACITANCE OF A PARALLEL PLATE CAPACITOR

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

Derivation of the Formula
As we known that electric intensity due to two charge sheets is
E = σ / Єo
Since charge density is given by
σ = Q / A
Substituting the value in above equation
E = Q / A Єo ——— (I)
According to the definition of electrical potential.
ΔV = E Δ
Substituting the value of E from eq (I) in above equation
Q = C x Qd / A Єo
C = A Єo / d
If some other medium is used instead of air then,
C = A Єo Єr / d

Find the formula for equivalent capacitance when,
1. Capacitors are connected in Series
2. Capacitors are connected in Parallel

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

WHEN CAPACITORS ARE CONNECTED IN SERIES OR SERIES COMBINATION

Consideration
Consider three capacitors having capacitance C1, C2, C3 connected in a series. These capacitors can be replace by an equivalent capacitor having capacitance Ce. 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 Vab, Vbc, Vcd.

Derivation of the Formula
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.
Vad = Vab + Vbc + Vcd ———— (I)
We know that,
Q = CV
V = Q / C
Therfore,
Vad = Q / Ce
Vab = Q / C1
Vbc = Q / C2
Vcd = Q / C3
Substituting the values of Vad, Vab, Vbc and Vcd in eq (I)
Q/Ce = Q/C1 + Q/C2 + Q/C3
Q/Ce = Q [1/C1 + 1/C2 + 1/C3]
1/Ce = 1/C1 + 1/C2 + 1/C3

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

WHEN CAPACITORS ARE CONNECTED IN PARALLEL OR PARALLEL COMBINATION

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

Derivation of the Formula
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 = Ce Vab
=> Q1 = C1 Vab
=> Q2 = C2 Vab
=> Q3 = C3 Vab
Substituting the values of Q1, Q2, Q3 in eq (I)
Ce Vab = C1 Vab + C2 Vab + C3 Vab
Ce Vab + Vab (C1 + C2 + C3)
Ce = C1 + C2 + C3

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

Qs. Define electric potential and absolute potential.

ELECTRIC POTENTIAL

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

Explanation
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.

Mathematical Form
Electric Potential is denoted by V. It is defined as the potential energy per unit charge. Therefore mathematically,
ΔV = Δu / q ——- (I)
We know that
Δu = F. Δr
But E = F/q => F = Eq, therefore,
Δu = EqΔr
Substituting the value in eq (I)
ΔV = EqΔr / q
=> ΔV = E. Δr
Where,
ΔV = Electric Potential
E = Electric Intensity
Δr – Displacement
From above expression we know that
Electric Potential is the dot product of electric intensity and displacement.
It is a scalar quantity and its unit is Volt.

ABSOLUTE POTENTIAL

Definition
Absolute potential is the amount of work done in order to bring a unit positive charge from one point to infinite distance against the direction of magnetic field.

ELECTRIC POTENTIAL
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.
Consideration
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. This patch is denoted by Δr.

Electrostatics MCQs with answers

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