Unit 2 Electrostatics-1 1

7/27/2019 Unit 2 Electrostatics-1 1

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UNIT 2 : ELECTROSTATICSInstructor: Shaikh Azeemuddin


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The sources of EMF waves are electric charges and strength of a field at any point

depends upon magnitude, position, velocity and acceleration of the charges.

Electrostatic field is a special type of EMFs, where charges are stationary and we

need only magnitude and direction of the charges.

1. Coulombs Law2. The Electric Field3. Gauss Law


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The sources of electromagnetic fields are electric charges

the strength of a field at any point depends upon themagnitude, position, velocity and acceleration of the

charges involved. In an electrostatic field sources are stationary, so that

only the magnitude and position of the charges need tobe considered..

3


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The force of attraction or repulsion betweenany two point charges is directly proportionalto the product of two charges and inverselyproportional to the square of distance

between them.

The force is repulsive when both charges arelike and attractive when the charges are

unlike


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Consider the two point charges Q1 and Q2separated by a distance r.

The force of interaction between twopoint charges is given as follows:

= Newton

Where K is the constant of proportionalityand depends on nature of medium

between the charges.


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The value of K is determined as

= 14 = 9 1 0 As force is a vector quantity so

=

Here is unit vector in the direction in which

the force is applied.


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is the permittivity of the medium ordielectric constant which is written as

= Farad/metre. Relative permittivity of the medium = Permittivity of the free space or vacuum,

= 36 =8.854102 F/m


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Example 2.1

Find the force on charge Q1, 20 C due to charge Q2, - 300 C where Q1 is at (0, 1, 2) m an

Q2 at (2, 0, 0) m.

Example 2.2

Point charge Q1= 300 C located at (-1, 1, -3) m experiences a force 1 = 8 8 + 4

due to point charge Q2 at (3, -3, -2) m. Determine Q2.


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=

So electric field,

= = N/C

Example 2.3

Find electric field intensity, Eat (0, 3, 4) m due to a point charge, Q = 0.5 C at the

origin.


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If a test charge is moved towards the charge Q, the

test charge will experience force due to the maincharge Q.

The lines of force is called as electric flux which isequal to the charge itself.

The electric flux from electric charge Q is given by =

Example 2.4

Three point charges, Q1 = 30 nC, Q2 = 150 nC and Q3 = - 70 nC are enclosed by surface S.What net flux crosses S?


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Electric flux per unit area.D = = =Coulomb/metreFor spherical surface area;

= = =

But =/Therefore = =DExample 2.5A point charge, Q = 80 nC, is located at the origin. Find the electric flux density D at (1,

0, -4).

Example 2.6

Find the flux density due to a charge of 3C positioned at (2, 2, 1) exerting a force on anothe

charge of 2C positioned at (-3, 0,-1).


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Surface Charge Density: The total charge distributed over a surface. This gives the total charge per unit area. It is given by

= Coulomb/sq metre (C/m)

Volume Charge Density: The total charge distributed over a volume.

This gives the total charge per unit volume. It is given by

= Coulomb/Cubic metre (C/m)

2.4 Continuous Charge Distribution and Gausss Law

If charges are distributed instead of concentrated at one point, it is better to define

charge distribution in terms of charge density. The different types of charge density are given

below.


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Gausss law is means of finding E or D for symmetricalcharge distribution.

The electric flux passing through any closed

surface is equal to the total charge enclosed by the

surface.


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The total flux passing through the closed surface is given by

totalQ

And electric flux density is defined as

Sd

d

SD S

where S is the surface area

SdDdSd

dD SS .

Or

)(. ASdD SS

We know that totalQ

In terms of volume charge density, Q can be obtained as:

)(BdvQ

dvdQ

dv

dQ

v

Q

v

v

v

Now using divergence theorem for equation (A).

v

S

S

S

CdvDSdD )().(.

Comparing both equations (B) and (C) we come to know that the divergence of electric flux

density is equal to the volume charge density.

)(. DDSv

Equations (A) and (D) are expressions for Gauss Law of electric field.


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Example 2.7

Find the charge in the volume defined by 0 1 , 0 1 and 0 1 , i

= 2 /

3

.What change occurs for the limits 1 0 ?


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Consider a uniform electric field E and test charge q, in

the field E.

If the test charge is moved from one point to another

point in the electric field E, a force acts on the test charge

due to the electric field.

This force is given by:

=

Since there is a movement of charge in the electric fieldfrom one point r1, to another point r2, there will be work

done against the force.


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The work done in moving Q a distance of dr is: = Joules

Integrating both sides,

= 2 Joules

Example 2.8

Calculate the work done in moving 2C charge from B(2,2,2) to A(3,-5,2).

The electric field is define as:

yx ayaxE22 420 mV /


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work done in moving a unit positive charge fromone point to another in an electric field

= =

= Volts

= = 2 =

Volts

This is the potential difference between two points, r1 and r2.

=

42

41

V = V2V1

If the test charge is moved from infinity to a given point in the electric field,

V1 = 0 1

1= 0

Then V = V2


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The work done in moving a unit positive charge frominfinity to a given point in an electric field is calledabsolute potential.

= Volts

1 = 0 = Example 2.9

A force NewtonsaaaF zyx 338 is applied on a charge of 1nC to move it over a

distance metersaaad zyx 7210 .

i. Find the vectors indicating direction of force and distance; andi i . Find the work done by this force.i i i . Find the absolute potential of this charge.


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The work done by an external force in moving a unit positive

charge from one point to the other is

= =

= Volts Consider the scalar potential V is a function of x, y, z ,then

above equation can be written as

+ + = This can be re-written as

+

+

. + + =

=

= Comparing both sides

VVE .

Thus the electric field strength at any point is just the negative of the potential gradient at that

point.


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Example 2.10

Given the potential field, = 62 + 6 + 4 volts and a point P (-4, 3, 6), find:

a. Potential V at point Pb. Electric field intensity E at point Pc. Electric flux density D at point Pd. Volume charge density at point P


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2 = is Poissons EquationPoissons equation for Cartesian co-ordinate system is written as

2 = 2

2 +

2

2 +

2

2 =

If the volume charge density is zero, then2 = 0 This is Laplaces equation.The operator 2 is called Laplacian operator.

According to Gausss law in point form, the divergence of electric flux density is

equal to the volume charge density.

. =

But, =

. =

. =

. =

E =

. =

. =

Unit 2 Electrostatics-1 1

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