electric_intensity.pdf

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Yangchenphug Higher Secondary SchoolDepartment of PhysicsPhysics: Class-12 Free e-book

------------------------------------------------------------------------------------------------------------www.royphysics.netau.net Mail [email protected]

BHSEC-I SC-CBSE sudipta kumar roy

Uni t : 1 El ect rostat i c

Topi c/ chapt er : El ect r i c di pol eSyl l abus: Electric dipole; electric field at a point on the axis

and perpendicular bisector of a dipole; electric dipole

moment, torque on a dipole in a uniform electric field .

[Scope: Electric dipole and dipole moment (vector);unit; derivation of E at any point, (a) on the

Axis (b) on the perpendicular bisector of the dipole.

Vector equation; special case for r>> 2l. Dipole in

uniform E field; net force zero, torque E p = . (E due

to continuous distribution of charge, ring of charge,

disc of charge, etc. not included).]LessonDevel opment : El ect r i c Fi el d

The space surrounding an electric charge q in whichanother charge qo experiences a force of attraction orrepulsion, is called electric field of charge q.

The intensity of electric field E is given by

o

FE

q=

Electric intensity is a vector quantity. S.I unit of E is newton/coulomb or N/C. Its another unit is

Volts/metre [V/m].

El ect r i c di pol e

An electric dipole is a pair of equal and opposite

point-charges placed at a short distance apart.

-q +q

l l

The length of dipole is 2l.

El ect r i c di pol e moment

The product of one charge q and distance between dipole

2l is called magnitude of the electric dipole moment[p].

P = q x 2lPh

sicsnotesb

:S.

K.

Ro

forro

h

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Electric dipole moment is a vect or quant i t y which acts

along the line joining the two charges. Unit ofelectric dipole moment is coulomb-metre [C-m].

E f or el ect r i c di pol e on axi al l i ne :[E-electric intensity]

Let a test charge qo [It is a unit positive charge] beplaced at a distance of r from centre of dipole. Theelectric intensity due to positive charge is E1 and due

to negative charge is E2.The resultant intensity is:E = E1 E2 [As E1E2]

Or we can write,

12

1E

4 ( )o

q

r l=

ur

and 22

1E

4 ( )o

q

r l=

+

ur

Hence,

2 2

1 1E

( ) 4 ( )o o

q q

r l r l =

+

ur

Taking common factors out we get,

2

1 1E

4 ( ) ( )o

q

r l r l

= +

ur

We take L.C.M for ( r l )2 and (r + l)2

which gives us (r2 l2)2 and on solving we get:

2 2

2 2 2

( ) ( )E

4 ( )o

q r l r lr l

+ =

ur

And on further solving we get:

2 2 2

4E

4 ( )o

q rl

r l

=

ur

..equation 1

Now we know that electric dipole moment is given as:

p = q x 2l .equation 2

Physicsnotesby:S.K

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F2Cos

Put value of P from equation 2 in equation 1we get

2 2 2

1 2E

4 ( )o

pr

r l

=

ur

If we consider that r l[ negl ect l] then we get theelectric intensity as:

2 2

1 2E

4 ( )o

pr

r

=

ur

which can further be modified by canceling

r as:

31 2E

4

=o

pr

uur

E f or el ect r i c di pol e on equat or i al l i ne[E-electric intensity]

Also E = Fqo and since qo=1C hence E = F eq 2

We know that1 2

1

4 ( )o

qF

r l=

+and

2 2

1

4 ( )o

qF

r l=

+also

2 2Cos=

l

r l+

Now we can modify equation 1 as:

2 2

1 1F =

4 ( ) 4 ( )

+

+ +o o

q qCos Cos

r l r l

-q +q

F1

F2

F1 Sin

F1 Cos

F2 Sin

l l

We know that forces of

attraction or

repulsion are equal

and opposite of each

other.

As the factorsF1 Si n and F2 Si n are equal but opposite

to each other hence

they get cancelled.

Now the resultant

force acting on test

charge qo is:

1 2F = F FCos Cos + eq 1

qo [+1C]

Phys

icsnotesby:S.

K.

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Or,2

1F = 2

4 ( )

+o

qCos

r l

Or,2 2 2

1F = 2

4 ( ) + +o

q l

r l r l

Or,3

2

1F = 2

4 ( ) +o

ql

r l

Or,3

2

1F =

4 ( ) +o

p

r l

Now if rl then we get force as:[F-vector quantity]

3

1F =

4op

r

Tor que exper i enced by el ect r i c di pol e

F=qE

F=qE

2l

Torque is defined as product of

force and perpendicular distance

between forces.

force distancear =

Force = qE and perpendiculardistance is BC

We can writeBC

Sin2l

=

qE

qE 2

BC

lSin

= =

We also know that dipole moment

is given as: p = q x 2l

pE Sin =

This equation is scalar form and

the same equation can be written

in vector form as:

p x E = All three quantities are vector.

BC

A

Physic

snotesby:S.

K.

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Act i vi t y:

1. The electric field intensity at a large distance x isproportional to :

1/x2 1/x3 1/x 1/x4Dont

Know

2. Regarding the adjacent figure which statement is true:

EA EB

EA> E

BEA< E

BEA= E

B

None of

these

Dont

Know

3. If a test charge qo is placed in an uniform electric E

field due to a charge q then force experienced by test

charge is

F = qE F = 2qoE F = 2qE F = qoE

4. If an electric dipole consists of two charges 200C

separated by a distance of 10cm then calculate the electric

dipole moment.

5. If electric intensity on equatorial line is 3x10-2 N/C then

what would be the electric intensity on equatorial line?

6. What is the dimension of electric field intensity?2

3F MLTE MLT Aq AT

= = =

BBE- Quest i on

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