Electric Fields, Potential, and Capacitance: A Comprehensive Guide, Study notes of Physics

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CHAPTER

ELECTROSTAT

ICS

INTRODUCTION

Conservative force :-In case of conservative force , when the work is done against

the force ,

then that amount of work is stored in the form of potential energy

For example :- gravitational force , electrostatic force.

Gauss’ law :- The flux of the net electric field through a closed surface is equal

to the net charge enclosed by that surface divided by 𝜀 0

Ԧ

𝑄

0

It is the relation betweenthe electric charge and its electric field.

Q is the total charge enclosed by that surface.

𝜙 is the flux coming out of a closed surface.

Since , q = 𝜎 x

4 𝜋 R

2

2

We get E =

2

2

0

0

. Hence , E =. Thus the electric field at a point

outside

2

the shell

is

same as that due a point charge.

Hence , it can be concluded that a uniformly charged sphere is

equivalent to a point charge at its centre.

CASE I : If point P lies on the surface of sphere , then r = R ,

0

0

hence , E = =.

2

CASE II : If point P lies inside the sphere , then 𝜎 = 0 , since there is

no charge

inside the sphere.

Hence E = 0.

Consider a uniformly charged wire of infinite length with a

constant linear charge density 𝜆 ( charge per unit length ) ,

placed in a medium of

permittivity 𝜀 =

0

k.

Due to symmetry , magnitude of electric intensity is same at all

points on Gaussian cylinder and directed radially outward.

At point P , angle between 𝑬 and 𝒅𝒔 is zero. So cos 𝜃 = 1.

Hence , 𝑬 ∙ 𝒅𝒔 = E ds cos 𝜃 = E ds. Thus the flux d∅ through

area ds is = E ds.

Total electric flux through the Gaussian surface ∅ = ׯ 𝑬 ∙ 𝒅𝒔

= ׯ E ds

𝗌 0

∅ = E ׯ ds = E 2 𝜋 rl. But , ∅ 𝗌=

𝑞

,

ELECTRIC FIELD INTENSITY DUE TO AN

INFINITELY LONG ,STRAIGHT

CHARGED WIRE

ׯ

𝐀

Since 𝜆 = , or q

= 𝜆l

𝑙 ∴

𝜆

𝑙

𝗌 0

= E 2

𝜋r l

Hence , E

=

ׯ

𝐀 2𝜋𝗌 0

The direction of 𝑟 𝑬

isradially outward if 𝜆

is positive and

inward if 𝜆 is

l

+

+

+

+

+

+

+

d +

s

R

To find electric intensity at point P , at a distance r from the charged

wire ,

𝒅𝒔 P

consider a coaxial Gaussian cylinder of length l and radius r

passing through point P.

ds is the area surrounding P on Gaussian cylinder.

+

+

r

_ _

Gaussian

surface

EXAMPLE

8.

EXAMPLE

8.

GIVEN : R =

10cm Or R =

10 x

• m

q = 1 𝜇 C = 1x

• C ,

Calculate E (i) at r

= 30 cm

= 30 x 10

• m

from center ,

(ii) at surface i e.

r = R ,

(iii) at 5 cm from

center.

SOLUTION :

(i) E=

q/4𝜋𝜀

0

r

2

(ii) E = q / 4 𝜋𝜀

R

2

0

(iii) at 5 cm from

center means the

point is inside the

sphere.

i e. r <

R, Hence

E = 0.

EXAMPLE 8.

GIVEN :

Length of straight

thin wire

L = 2m ,

Q = 3 𝜇C =3 x

• 6

C ,

Calculate :

( i ) Charge

density 𝜆,

( ii ) Intens ity

at a point r

= 1.5 m ,

S O L U T I O N

( i ) 𝜆 =Q /L

Take r = 1.

m.

GIVEN : Charge per

unit area of a large

sheet 𝜎 = 3μC/m

2

𝜎 = 3 x 10

C/m

2

Find intensity near

the sheet

SOLUTIO

N :

E = 𝜎 /

0

Take 𝜀

0

= 8.85 x 10

12

SI units.

ELECTRIC POTENTIAL AND

POTENTIAL ENERGY

Potential energyof a system is the stored energy that

depends upon the relative positions of the charges in that

system.

It is the work done against the electrostatic forces to gain

certain configuration of charges.

Every system tries to remain in lowest potential energy level.

Always there is some force between the charges .Either attractive

or repulsive.

Hence always there is some work done whenever one charge is

moved in the vicinity of the other.

Therefore potential energy arises from any collection of charges.

The work done in bringing any charge 𝑄 in the vicinity of other

charge is equal to change in potential energy of this system.

The change in P. E. depends only upon the end points and is

independent of the actual path taken by the charge to move from one

point to the other.

This change in P. E. is equal to the work done W

AB

against the

electrostatic force.

W

AB

= ∆ U

0

2

1

Q q

0

[ -

]

It is convenient to choose infinity to be the point of zero P. E. ,

since the electrostatic force is zero at r = ∞.

Hence the potential energy U of the system

consisting of two point charges q

1

and q

2

separated by r is

obtained using r

1

= ∞ and r

2

= r.

It is given by , U (r) = [

] [

1

2

]

4𝜋 𝗌 0

𝑟

UNITS OF POTENTIAL

ENERGY

 SI unit of potential energy is joule.

 One joule of energy is that that energy which is stored in

moving a charge of one coulomb through the potential

difference of 1 volt.

 Electron volt is the change in kinetic energy of an

electron while crossing a potential difference of 1volt.

 1eV = 1.6 x 10

• joule

 1meV = 1.6 x 10

• x 10 - = 1.6 x 10 - joule

 1keV = 1.6 x 10

• x 10

3 = 1.6 x 10

• joule

RELATION BETWEEN ELECTRIC FIELD AND ELECTRIC

POTENTIAL

Consider a charge +Q at the given point.

If one coulomb of charge is moved through the distance of dx

in the field of this charge , then the work done is ,

dW = - 𝑭

. dx= - Edx here q = 1 C

Hence dW = dV = - E dx

E = - dV / dx

Therefore electric field is the negative potential gradient at

the given point

ZERO POTENTIAL

Every potential is measured with respect to the zero potential

For a point charge or localized collection of charges , zero

potential is INFINITY.

THUS THE POTENTIAL AT A GIVEN POINT IN AN ELECTRIC FIELD IS

THE AMOUNT OF WORK DONE TO BRING A UNIT POSITIVE CHARGE

FROM INFINITY TO THE GIVEN POINT.

EXAMPLE

Potential^ 8.4 at point

A is V A

= 4 X 10

5

V.

i) Find work

done.

q = 3 μC, V A

=

W/q Or W = q

V A

= 3 X 10

• 6 X 4 X 10

5

J

EXAMPLE

W AB

= 120 J , q =

6 C , V A

= 10 V and

V B

= V , V =?

V B

• V A

=

W A B

/q V –

10 = 120 /

V = 20

• 10 =

30 V.

a) ELECTRIC POTENTIAL DUE

TO A POINT CHARGE

Consider a point charge +q at point O. To

find the potential at point A , at a distance r

from O.

The potential at point A is the amount of work

done per unit positive charge brought from ∞ to

point A.

We select a convenient path along the line

extending

OA to ∞ ( since the work done is independent of

path .) M is an intermediate point on the path and

OM = x.

The electrostatic force on a unit positive charge

at M is ,

2

F = ( for 1 coulomb of

charge ). 0

The force is directed away from O along OM.

For the infinitesimal displacement dx from M to N

, the amount of work done is ,

dW = - F dx ( displacement is opposite to the

force )

+q

O

A

r

dx

x

N^ M

The total work done in displacing a

unit

positive charge from ∞ to A is ,

∞

W =

𝑟

∞

𝑟

ׯ

ׯ

0

4𝜋 𝗌

𝑥

2

d

x

0

∞

𝑟

− 2

dx

ׯ

ׯ 4 𝜋

𝗌

0

ׯ

𝐀

[ -

1

]

𝐀

𝐀

− 2 dx = -

1

)

𝑞 1

1

[ -

]

1

∞

W

4𝜋𝗌 0

𝑟 ∞

𝒒

𝟒𝝅𝗌 𝟎

r

In this way , the electrostatic potential at point A due to charge q is ,

V = W =

0

A positively charged particle produces a positive electric potential and a

negatively charged

particle produces a negative potential.

At infinity , r = ∞ , V =

∞

This shows that the electrostatic potential at infinity is ZERO.

The electrostatic potential due to a single charge is SPHERICALLY SYMMETRIC. ׯ

𝐀

The variation of electric potential

V 𝛼

1

and

1

𝑟

2

𝑟

2

the variation of electric field E 𝛼

1

with

r , 1

ׯ

ׯ

the distance from the charge is shown in the diagram.

electric field (E)

electric potential

(V) O distance

r

V

E

By the geometry of the

diagram ,

r 1 = r + l – 2rlcos

𝜃 2 2 2

r 2 = r + l + 2rlcos

𝜃 2 2 2

1

2

2

Now , r = r [

𝐀

𝐀

2

𝑟

2

2

𝑙

cos 𝜃

,

and

] , and

2

2

2

2

𝑟

2

r = r [1 + +

2

𝑙 𝑙

cos 𝜃 𝑟

𝑟

]

𝐀

𝐀

For a short electric dipole , 2l <<< r or if r >>> l

, then

𝑙

is small

an

d

𝑙

2

𝑟

2

can be

neglected.

1

2

2

So , r = r [ 1 –

2

𝑙

cos 𝜃

an

d

2

2

2

r = r [ 1 +

2

𝑙

cos 𝜃

]

∴ r 1

= r [ 1 –

𝑟

𝑙

cos 𝜃 𝐀

𝐀

] ,

1

]^2 similarly , r 2

= r [ 1

𝑟

𝑙

cos 𝜃 ׯ

ׯ

1

]^2

r

r 1

r 2

l

l

O

ׯ

ׯ

ׯ

ׯ

C

Hence we can

write ,

(^1) 𝐀

𝐀

1

ׯ

ׯ

1 [ 1

𝐀

𝐀

2𝑙 cos 𝜃

]

−1/

an

d

1

ׯ

ׯ

2

ׯ

ׯ

1 [ 1

𝐀

𝐀

2𝑙 cos 𝜃

]

−1/ c 1

2

∴ V = V+ V

=

𝑞

4 𝜋

𝗌

0

1

𝐀

𝐀

[ [ 1

𝐀

𝐀

2 𝑙 cos 𝜃 ]

− 1 / 2

1

[ 1 +

𝑟

𝑟

2𝑙 cos 𝜃

]

−1/ ] Using binomial

expansion ,

1 + 𝑥

𝑛 = 1 + n x ,

x << l , neglecting higher order terms ,

c

V

=

ׯ

𝐀

0

4 𝜋 𝗌

𝑟

1

[ ( 1

0

0

𝑙 cos 𝜃

) – ( 1 −

𝑙 cos 𝜃

) ] =

𝑞 1

[ 1+

𝑙 cos 𝜃

• 1 +

𝑙 cos 𝜃

] =

𝑞 1

[

2 𝑙 cos 𝜃

]

c

Hence , V

4 𝜋

𝗌

0

𝑟

1 𝑝 cos

𝜃

𝑟

2

𝑟 4 𝜋 𝗌 𝑟 𝑟 𝑟

4 𝜋 𝗌 𝑟 𝑟

( p = q X 2l ).



The electric potential at point C is also

expressed as ,

c

V

3

c

or V

0

0

2

( 𝑟Ƹ =

𝑟Ԧ

). 

Potential at an axial point , 𝜃 = 0 ° (towards +q ) or 180 °

towards –q ,

V

axial

=

0

2 

This is the maximum value of potential.



The potential at an equatorial point , 𝜃 = 90 ° , V = 0.



This is the minimum value of potential.



Hence the plane perpendicular to the line between the two charges

passing through the midpoint of the dipole is an equatorial plane with

ZERO potential



The work done to move the charge anywhere in an equatorial