Physics

ELECTROSTATICS - 3

0

q

Application of Gauss’ law

Electric potential (intro.)

Session Objectives

Gaussian surface

Any imaginary closed surface.

Gaussian surfaces are useful in computingElectric field (usually uniform) and flux

Gaussian surfaces are very useful for finding electric field with symmetrical charge distribution

Electric field due to an infinitely long straight charged wire

Due to cylindrical symmetry of charge, electric field is away from the centre of the wireThe net electric flux,

ˆ. . 0B C

E ds = E ds as E n

A A

E.ds = Eds =E.2 rh as E ll n ˆ

=E. 2 rh

A B C

= E ds + E ds + E ds

ˆE n

ˆE n

Charge enclosed by the cylinder, q = h

By Gauss’ law,0

q=

ε

0

λhE.2 rh=

ε

E

+++++++++++++

r

1i.e., E

r

0

λE=

2 ε r

Solved Example – 1

An infinite line charge produces a field of 9×104 N C–1 at a distance of 2 cm. Calculate the linear charge density.

Solution:

E = 9×104 N C–1, r = 2×10–2 m

4 –2–7 –1

9

1 9×10 ×2×10= × =10 Cm

9×10 2

0

1 λE=

2 ε r

0

Erλ=4 ε ×

2

Solved Example –2

An electric dipole consists of charges +1.6 nCand –1.6 nC separated by a distance of 2×10–3 m.It is placed near a long line charge of linear charge density 5×10–4 Cm–1 as shown in thefigure. If the negative charge is at a distance of2 cm from the line charge, then, find the net forceon the dipole

– +2 cm

Solution

= 5×10–4 Cm–1, r1 = 2×10–2 m, r2 =2.2×10–2 m,q1 =-1.6×10–9 C, q2 =1.6×10–9 C

Electric field due to a line charge at a distance r from it,

0 0

2E

2 r 4 r

Field at the point of negative charge, 9 –4

1 –20 1

2 2 9 10 5 10E

4 r 2 10

Force on the negative charge

9 –4 –9

1 1 –2

2 9 10 5 10 1.6 10F E q 0.72N

2 10

Solution Contd.

Electric field at the point of positive charge is

9 4

2 –20 2

2 2 9 10 5 10E

4 r 2.2 10

Force on the positive charge

9 –4 –9

2 2 –2

2 9 10 5 10 1.6 10F E q 0.65N

2.2 10

F1 is towards the line charge

F2 is away from the line charge

Net force on the dipole,

F =F1 – F2 = 0.72 – 0.65 =0.07 N, towards the line charge

– +

F1 F2

Solved Example - 3

A long cylindrical wire carries a positive chargeof linear charge density 4×10–7 Cm–1. An electron revolves round the wire in a circular pathunder the influence of the electrostatic force. Find the KE of the electron.

Solution

rLet the electron revolves in a circular pathof radius r. Electrostatic force on the electronprovides the necessary centripetal force.

Field at a distance r from the wire of charge,

0 0

2E

2 r 4 r

Solution Contd.

Force on the electron,0

2 eF Ee

4 r

Required centripetal force2

cmv

Fr

2

c0

mv 2 eF F

r 4 r

Kinetic energy of the electron,

2

0

9 –7 –19

–17

1 eKE mv

2 4

9 10 4 10 1.6 10

5.76 10 J

0

σE=

2ε

Electric field due to infinite plane sheet of charge

Thin, infinite, nonconducting sheet having uniform surface charge density, as a result of which surface of the sheetE

??????????????

E??????????????

E??????????????

At R and S, E ll n

Electric flux at plane faces,

=2E.ds =2Eds

Electric flux over the curved surface is zero, as no field lines crosses it

Total electric flux, =2Eds

Proof.

ˆE|| n

ˆE|| n

Charge enclosed by the Gaussian surface,

q=σ ds

By Gauss’ law,0

q=

ε

0 0

σ ds σ2E ds = E=

ε 2ε

E is independent of the distance from the plane sheet

A B

A B

0

(σ +σ )=–

2ε

A B

0

(σ – σ )=

2ε

Two infinite parallel sheets of charge

1 2E = – E + E??????????????????????????????????????????

1 2E = E – E??????????????????????????????????????????

Due to uniform surface charge density, surface of the sheetE ??????????????

Region II

Region I

A B

Two infinite parallel sheets of charge

Region III

A B

0

(σ +σ )=

2ε

1 2E = E + E??????????????????????????????????????????

The sheets have equal and opposite charge density

–

I II III

E 0 E 00

E

If and

, then

A

B –

In regions I and III, E = 0

In region II,

A Bas ( – ) 0

0 0

– (– )E

2

Special Case

Solved Example – 4

Two large thin metal plates with surface charge densitiesof opposite signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed parallel and close to each other. What is the field(i) To the left of the plates?(ii) To the right of the plates?(iii) Between the plates?

Solution:Electric field exists only in the region between the plates. Therefore,

(i), (ii) E = 0

–20 –2

–208 –1

–120

(iii) σ =44.27×10 Cm ,

σ 44.27×10E= = =5×10 NC

ε 8.854×10

Applying Gauss’ Law – Spherical Symmetry

0S

qE.ds =

ε

(i) Field at an outside point, i.e., r > R

22

0 0

q 1 qE×4 r = E=

ε 4 r

The field is the same as if the whole charge is placed at the centre of the shell

Note that E is same at all pointson the Gaussian surface

Electric field due a charged sphere of charge q and radius R

ˆE ||n??????????????

Gaussian surface

(ii) At a point on the surface,r = R

(iii) At a point inside the spherical shell r < R

20

1 qE=

4 ε R

No charge is enclosed by the Gaussian surface.

S

E.ds 0 E 0

Gaussian surface

r

q+

+

+ ++

+

+++

+

r

20

1 qE= for r R

4 ε r

E=0, for r <R

Variation of electric field intensity from the centre of the shellwith distance

Solved Example –5

The uniform surface charge density on a spherical copper shell is What is the electric field strength on the surface of the shell?

Solution:

o

20

1 qE=

4 ε R

The electric field on the surface of a uniformly charged spherical conductor is given by,

A spherical charged conductor has a uniform surface charge density . The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

Solved Example –6

Solution:

20

1 qE=

4 ε R

2

2o0

4 R

4 R

The electric field on the surface of a uniformly charged spherical conductor is given by,

Thus, E is independent of the radius of the sphere. As is constant, E remains the same.

Solved Example –7

Solution:

1

o

1

1 o

E 1

1E

The electric field on the surface of a charged spherical conductor is given by,

(i)

The uniform surface density of a spherical conductor is and the electric field on its surface is E1. The uniform surface density of an infinite cylindrical conductor is and the electric field on its surface is E2. Is the expression correct?

1

2

1 2 2 1E E

Solution Contd.

The electric field on the surface of a charged cylinder is given by,

( r = radius and = length )

(ii)

From equations (i) and (ii) we get,

1 2 2 1E E

2E

2

o o

q

2 r

2

2 o

E 1

Solved Example – 8

A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell(b) Just outside the shell(c) At a point 15 cm away from the centre

Solution:

q = 2 ×10–6 C, R = 0.1 m, r = 0.15 m

(a) Inside the shell, electric field is zero

–69

2 20

6 –1

1 q 2×10(b) E= =9×10 ×

4 ε R 0.1

=1.8×10 NC

–69

2 20

5 –1

1 q 2×10(c) E'= =9×10 ×

4 ε r 0.15

=8×10 NC

Solution Contd.

Electric potential

Electric potential at a point - the work done in bringing unit positive charge from infinity to that point against the electric forces.

WV =

q

SI unit – Volt (V)

1 J1 V =

1 C

One volt - the electric potential at a point if one joule of work is done in bringing unit positive charge from infinity to that point.

Solved Example - 9

If a positive charge be moved against the electric field, then what will happen to the energy of the system?

Solution:

If a positive charge be moved against the electric field, then energy will be used from an outside source.

Solved Example -10

If 80 J of work is required to transfer 4 C charge from infinity to a point, find the potential at that point

Solution:

W 80V = = =20 V

Q 4

W =80 J, q = 4 C, V =?

Thank You