Conductor 1

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CONDUCTOR AND ITS PROPERTIES 3. Conductor is an equipotential body. Every point on conductor will have same potential. F- inside

AV

Home work

TÉdem

0

-_

or

for

¥

any

points

two

metallic

(

( EE9E

÷¥o

All

point

on

surface

equipotential Electric field along surface are

must

cost

.

be

zero

-0

0=900

5. No electric field lines are present inside the conductor and electric field on surface is perpendicular to it. VJ

1010¥

.

chapter -1

Read

4. At the surface of a charged conductor, electrostatic field must be normal to surface at every point.

of bodies

conductor

NCERT -

.

except

and

Solve

After completing

exercise

module)

.

ILLUSTRATION

10¥ A neutral spherical conductor of radius R is kept near a

V=k¥

point charge as shown. a) Calculate potential of conductor b) Calculate potential at B due to induced charge c) What will be the magnitude and direction of electric field at centre due to induced charge?

¥

And

VB==Vq-Y =

VB Wind

✓ centre

=

VB

=

-

-

+

R

Vind -

+

'

r



+

q

-

+

Ed

(a)

Veenduotou

V=Vq

+

V=kfVJ

¥

End

-

-

B "

£

-_

to at

pt

.

due

charge

B

%¥=µ

=o

Éind Éq=o +

=

Ventre

Wind

*

-

④→

Potential

KILEY

Eq

=

k§z

Eind=k¥





AL

CONDUCTOR WITH CAVITY (ELECTROSTATIC SHIELDING) If there is no charge present inside the cavity then electric field inside cavity is zero. All the charge given to the conductor will lie on the outer surface of the conductor.

VJ

CAVITY WITH CHARGE Equal and opposite charge is induced on the inner surface of the conductor. Q

clout

point

: 2C

-

is

E. F.

surface

oh -19in ,

5C +

9. out

Gout =

=

Goat

=

=

=

=

-

=

°

0

-9 ,

Conserved

Q -9in @

surface



fending 9in

every

gaussian

of &

9in

VJ

at

zero

C- %)

0+9

=

Q

.

ILLUSTRATION A point charge q is kept inside the cavity inside a spherical conductor as shown. What will be the charge present on inner and outer surface of the conductor? +

-

q

q

q

q

(A)

VJ

B)

ILLUSTRATION Three spherical hollow conducting shells are given charge q, −2q and 4q respectively. Find the charge present on inner and outer surface of all three conductors. 4q − 2q q °

0

-

¥;

Says

"



"

"" ↳



,

52--5



£

% -1%2=9 ,

9s

VJ

--

,

99k

q

94=-9 9ns

,

-



-19,2+9%+94+955=0

§

Is > •

-

.

9%+94=-29

§ S

=

=

9s

=

-_

,

-19

39

As >

=

-94

1.

2.

Resultant electric field due to charge on S1 and inside S1 is zero on S2 and outside S2. Resultant electric field due to charge on S2 andfinside S2 b- S1. is zero on S1 and outside outside

inside

No

net

due to No net

S2

9,

S1

q1

q q2

VJ

,

force

an

charge

force 92

9,

ons , & q on

and

9 on

due to

52

52=+9

51=-9

DISTRIBUTION TYPE Sz

Sz

Sz

S,

s,

s

S

q

q

q

S1-7NU

,→U sz→U s

p

z

S

,

S,→NU

{



4

5

{





5



NU

{



U

,

U

Sz

9

,

q

q

5.

,

q

Sz

q

VJ

,

z

{→U

Sz s

Sz

U NU

6

4

NU

sz→NU

>

5

NU

{

NU

,

ILLUSTRATION An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge at centre. Calculate: a) electric field at point A and B b) Potential at point A and B cA=r, ^

52

CAB

µ Rz S,

p,

q

A

=

V2 VB



¥

→d¥

¥-2

%+¥s

=

'



ES2

=

Eps

If

>

¥?

__

EB

+ ,

Us

,

=k÷

Va=Vsz ( Surface E)



✓A

"

EA= F-

0

rs=Éq÷É+És

V,

=

-

,

0

VJ

=

1%-2

V1

+

"¥,

Us

,

-1

+

Use

:# ¥

"

+

ILLUSTRATION An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge at a distance of R1/2 as shown. Calculate: a) electric field at point A and B b) Potential at point A, B and centre 52

¥

F- A

=

0

E☐=%£ ¥

vB=*÷

S,

"

F

VA

t .

'

E

q



'

A

B

,

Ed

"

s

,



{→

VJ

NO



=

Us

,

,

V LR2

=

=

Vq Us +

,

+

Vsz

=¥÷;"k÷¥

,

U

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