Crackiitjee.in.Phy.ch25

CURRENT ELECTRICITY – ILLUSTRATIONS

Q. What would he the interaction force between two copper spheres, each of mass 1 g, separated by the distance 1 m, if the total electronic charge in them differed from the total charge of the nuclei by one percent?

Assume spheres to be point charges:

The atomic weight of copper is 63.55 g The number of atoms in 63.55 g

= 6.022 × 1023 atoms

 The number of atoms in 1 g of copper is N1 =

23 6.022 10

63.55 

The number of protons in 1 g of copper is n1 = N1 × atomic number of copper

= 29 N1

 Total charge on the nuclei of 1 g of copper atoms is q = n1e

= 29 N1e

According to problem, charge on each sphere is

q = q1 × 1% =

q

1

100

=

29N e

1

100

= 23 29e 6.022 10 100 63.55    F = 2 2 0

q

4



r

=

 

2 23 9 2 29e 6.022 10 9 10 63.55 1      _{ }   Here, e = 1.6 × 10–19 C

Substitute the value of e, we get F = 1.74 × 1015 N

Q. Calculate the ratio of the electrostatic to gravitational interaction forces between two electrons, between two protons. At what value of the specific charge

q

m

of a particle would these forces become equal

(in their absolute values) in the case of interaction of identical particles?

Gravitational force of interaction is given by

Fg = 1₂ 2

Gm m

r

(In magnitude)

The force of electrostatic interaction between two point charges is given by Fe = 1 2₂ 0 q q 4 r (In magnitude) For proton, q1 = q2 = + e = 1.6 × 10–19 C, m1 = m2 = m = 1.67 × 10–27 kg  Ratio = e g

F

F

= 2 2 0 2 2 e 4 r Gm r  = 2 2 0

e

4



m G

Here,

crackiitjee.in

G = 6.67 × 10–11 Nm2/kg2, 9 0 1 9 10 4   Nm 2_/C2

Putting the values, ratio for proton is

42 e g

F

_{4 10}

F





For electron, m1 = m2 = m = 9.1 × 10–31 kg, q1 = q2 = e = 1.6 × 10–19 C  Ratio = 1036

For being both forces are equal in magnitude, Fe = Fg or 2 2 2 2 0

q

Gm

4



r



r

or 2 0 2 q 4 G m   or

q

4

₀

G

m





…(i) Here,

crackiitjee.in

0 = 8.85 × 10–12 C2/Nm2

Substituting the values, in equation (i)

q

m



0.86 × 10

–10 _C/kg

Q. Two positive charges q1 and q2 are located at the points with radius

vectors r1 and r2. Find a negative charge q3 and a radius vector r3 of the

point at which it has to be placed for the force acting on each of the three charges to be equal to zero.

Since the electric force on q1 due to q2 is





1 2 1 2 12 3 0 1 2 q q r r F 4 r r     Similarly,





1 3 1 3 13 3 0 1 3 q q r r F 4 r r     and ₂₃ 2 3



2 3



₃ 0 2 3 q q r r F 4 r r     For equilibrium of q3, 31 32

F



F



0

or 3 1



3 1



₃ 3 2



3 2



₃ 0 3 1 0 3 2 q q r r q q r r 0 4 r r 4 r r         …(i)

crackiitjee.in

For equilibrium of q3, the unit vector along.

F

31 should be in opposite

direction of unit vector along

F

₃₂

i.e., 31 32 31 32 F F F F   But F || r₃₁



₃ r₁



So, unit vector along

F

₃₁ will be same as unit vector along



r

₃



r

₁





3 1



31 3 1 31 r r F r r F     Also, F || r₃₂



₃ r₂



32 3 2 3 2 32 F r r r r F     But 32 31 32 31 F F F   F or 3 2



3 1



3 2 3 1 r r r r r r r r       …(ii)

From Eq. (i) and (ii), we get









3 1 3 1 2 3 3 2 3 3 0 3 1 0 3 2 q q r r q q r r 0 4 r r 4 r r        

crackiitjee.in

or 1 3 3 ₂ 1 2 3 ₂ 3 2 3 2 0 3 2 3 1 0 3 2 q q r r q q r r r r 4 r r r r 4 r r           or 1 ₂ 2 ₂ 3 2 3 2 q q r r  r r







3 2



3 1 3 1 3 2

r

r

r

r

r

r

r

r





_







_

_







or 1 2 3 2 3 2 q q r r  r r or

q r

_{1 3}



r

₂



q r

_{2 3}



r

₂ or

q r

₁



₃



r

₂





q

₂





r

₃



r

₂



or r₃



q₁  q₂



 q r_{1 2}  q r_{2 1} 1 2 2 1 3 1 2

q r

q r

r

q

q









…(iii)

Similarly for equilibrium for q1,

3 2 2 3 1 2 3

q r q r

r

q

q





…(iv)

On putting the value of r₃ from equ. (iii) in Eq. (iv), we get

3 2 2 3 1 2 3

q r

q r

r

q

q



 



crackiitjee.in

or r₁



q₂  q₃



 q r_{3 2}  q r_{2 3} or r₁



q₂  q₃



 q r_{3 2}  q₂



1 12 22



1

q r

q r

q

q



_





















or r₁



q₂  q₃



 q r_{3 2}  q₂



1 12 22



1

q r

q r

q

q



_





















or r₁



q₂  q₃



=













3 1 2 2 2 1 2 2 2 1 2

q

q

q r

q

q r

q r

q

q









or r₁



q₂  q₃



q₁  q₂



= q₃



q₁  q r₂



₂  q₂



q r_{1 2}  q r_{2 1}



After solving, we get

q3 =



1 2



2 1 2

q q

q

q





and from Eq. (iii), we obtain

1 2 2 1 3 1 2

r q

r q

r

q

q







Q. A thin wire ring of radius r has an electric charge q. What will be the increment of the force stretching the wire if a point charge q0 is placed

at the ring's centre?

Because of force of electrical interaction, ring has a tendency to expand. So, a force of tension comes into play which balances the electrical forces acted on different element of ring.

For calculation of force of tension, we consider an element l on the circumference of ring.

The charge on considered element is

q = ll

where,

q

2 r

 



The force of interaction between considered element on ring and charge q0 is. Fe = 0 ₂ 0 q q 4 r   = 0 ₂ 0 q l 4 r  

crackiitjee.in

Here, l r    0 e 0 q F 4 r     e

2T sin

F

2



_

or 0 0 q T 4 r     0 0 q T 4 r    

Substituting the value of

q

,

2 r

 



0 2 2 0 q q T 8 r   

Q. A thin half-ring of radius R = 20 cm is uniformly charged with a total charge q = 0.70 nC. Find the magnitude of the electric field strength at the curvature centre of this half-ring.

Let us assume circular arc of radius R making an angle  at the centre of curvature of the arc. The arc is uniformly charged with charge q.

An element of charge dq is considered on the arc. Axis yy' is bisector of angle .

The electric field at point O due to considered element is

dE = ₂ 0 dq 4 R

q

dq



d





Also,

dE = – dE cos 

ˆj

– dE sin 

ˆi

= _{2 2} 0 0 dq _{cos jˆ} dq _{sin iˆ} 4 R 4 R  _{  }   = _{2 2} 0 0 q _{cos d jˆ} q _{sin d iˆ} 4 R 4 R  _{    }    

 The component of electric field along y-axis is Ey = /2 2 0 /2

q

cos d

4

R

 



 







= ₂





/2_/2 0 q _sin 4 R    _   or Ey = ₂ 0

q

_sin

_sin

4

R

2

2







_











_{ }

_

crackiitjee.in

= ₂ 0 q sin 2 R 2     Also, Ex = /2 /2

dEsin

 







= /2 2 0 /2

dq

sin

4

R

 









= /2 2 0 /2

q

sin d

4

R

 



 

 



= ₂





/2_/2 0 q _cos 4 R    _{ }    Ex = 0

 Net electric field is E =

E

2_x



E

2_y 2 0 2 E sin 2 R 2    

(along bisector of angle ) For semi-circular ring,

  

2 0 q E sin 2 R 2    2 2 0 q E sin 2 R 2      = _{2 2} 0 q 2  R

Substituting the values, we get E = 0.1 kV/m

Q. A point charge q is located at the centre of a thin ring of radius R with uniformly distributed charge –q. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance x from its centre, if x >> R.

This problem is based upon superposition principle of electric field.

1 2

E E





E

Here,

E = resultant electric field,

1

E

= electric field due to ring.

2

E

= electric field due to point charge on the centre of ring. According to superposition principle

E and E

_{1 2} are physically independent to each other.

The electric field at point P due to point charge q is 2 2 0 q _ˆ E i 4 x  

(along axis of ring)

Electric field at point P due to ring is





1 _{2 2} 3 /2 0

qx

_ˆ

E

i

4

R

x









1 2

E E

E

 



=





3 /2 2 _{2 2} 0

q

1

x

_ˆi

4

x

_R

_x







_









_





or









3 /2 2 2 3 3 /2 2 2 2 0

R

x

x

q

_ˆ

E

i

4

_{x R}

_x



_

_















_





=





3 /2 2 3 3 2 3 /2 2 2 2 0

R

x 1

x

x

q

4

_{x R}

_x



_

_





_



_



















_













or





2 3 2 0 3 /2 2 2 2

q

3R

x 1

....

4

2x

E

x R

x















_

_





_





_





crackiitjee.in

Since, x >> R, so in binomial expansion higher power may be neglected.





2 3 /2 2 2 2 0

3R x

i

2

q

E

4

_{x R}

_x













 



_

But R2 may be neglected with respect to x2.

2 5 0

3qR x

E

8

x

 



= 2 4 0

3qR

8



x

Q. A thin non-conducting ring of radius R has a linear charge density  =  0

cos , where0 is a constant,  is the azimuthal angle. Find the

magnitude of the electric field strength: (a) at the centre of the ring;

(b) on the axis of the ring as a function of the distance x from its centre. Investigate the obtained function at x >> R.

Here, the meaning of azimuthal angle is the angle with diameter. Since,  is cosine function of . So, ring if not uniformly charged. A part of ring is positively charged and remaining part of ring is negatively charged. The distribution of charge on ring.

0cos    

So, first and fourth quadrants are positively charged but second and third quadrants are negatively charged.

Now we consider a small element dx = Rd  on the ring. The charge on considered element is

dq = dx = 0 cos  Rd 

= 0 R cos  d

 Net electric field at point P is. E = E1 + E2 =





2 0 3 /2 2 2 0 R 4 R x   

when x >> R, R2 may be neglected with respect to x2.

 E = 2 0 3 0

R

4 x





when x >> R = 2 0 3 0

R

4

x

 



= ₃ 0 p 4 x where, p =  ₀ R2

crackiitjee.in

(a) For electric field at the centre of ring, x = 0,





2 0 3 /2 2 2 0 R E 4 R x     At centre, x = 0 2 0 3 0

R

E

4 R



 



= 0 0 4 R  

(b) Let the ring plane coincides with y-z plane. We consider a small element AB (of length dl) on ring.

Here dl = RdQ

where R is the radius of ring. y = R sin 

and z = R cos 

The electric charge on the considered element is dq = dl = 0cos  (R d)

= 0R cos d

The axis of the ring is X-axis.

The electric field at point P due to considered element is



p A



3 0 p A

dq r

r

dE

4

r

r









or



0





₃



0 ˆ ˆ ˆ R cos d xi yj zk dE ˆ ˆ ˆ 4 xi yj zk          or









0 3 /2 2 2 2 0

ˆ

ˆ

ˆ

xi R sin j R cos k

R cos d

dE

4

_x

_y

_z



 





 





_

_

=









0 3 /2 2 2 2 2 2 0

ˆ

ˆ

ˆ

xi R sin j R cos k

R cos d

4

_x

_{R sin}

_{R cos}



 





 



_

_{ }

_

=









0 3 /2 2 2 0

ˆ

ˆ

ˆ

xi R sin j R cos k

R cos d

4

_x

_R



 





 



_



₂0 ₂



3 /2 0

R

dE

4

R

x











2 2 ˆ ˆ

x cos d i R sin cos d j ˆ R cos R cos d k  _{     }         





0 x _{2 2} 3 /2 0

Rx cos d

dE

4

R

x



 









crackiitjee.in

dEy =





2 0 3 /2 2 2 0 R sin cos d 4 R x        and dEz =





2 2 0 3 /2 2 2 0 R cos d 4 R x      x x E dE  



=





2 0 3 /2 2 2 0 0

Rx

cos d

4

R

x





 







After integrating, Ex = 0 and Ey =



dEy =





2 2 0 3 /2 2 2 0 0

R

sin2

_d

2

4

R

x







_







=





2 2 0 3 /2 2 2 0 0

R

cos2

₀

2

8

R

x











_{ }













y

E

0





Similarly, Ez =



dEz

crackiitjee.in

=





2 2 2 0 3 /2 2 2 0 0

R

cos

d

4

R

x





 







=





2 0 3 /2 2 2 0 R 4 R x    xˆ yˆ zˆ E E i E j E k     z

ˆ

E E

E k









Ex  0, Ey  0



=





2 0 3 /2 2 2 0 R 4 R x    For x >> R, R2 + x2 x2 2 0 3 0

R

E

4 x



 



Q. A very long straight uniformly charged thread carries a charge  per unit length. Find the magnitude and direction of the electric field strength at a point which is at a distance y from the thread and lies on the perpendicular passing through one of the thread's ends.

Since we know that

E|| =



1 2



0 cos cos 4 r     



1 2



0 E sin sin 4 r       

The net electric field is E =

E

2_



E

_||2

The situation of problem is Here,

1 = 0°

Since, two parallel lines meets at infinity.

2

₂



  

and r = y 0

E

sin0 sin

4

y

2













_



_



_

_

|| 0

E

cos0 cos

4

y

2













_



_



_

_

= 0 4 y  

The net electric field is

E =

E

2_



E

_||2 or E = 0 2 4 y   Also, tan  = ||

E

1

E





45    

It means electric field E is directed at the angle 45° to the thread.

Q. A thread carrying a uniform charge  per unit length has the

conurations. Assuming a curvature, radius R to be considerably less than the length of the thread, find the magnitude of the electric field strength at the point O.

This problem is based upon superposition principle of field for continuous distribution of charge.

1 2 3

E E





E



E

(a) The thread of the problem may be divided into three parts. (i) Semi-infinite vertical thin thread.

 We know that 0 E 4 R    

crackiitjee.in

|| 0 E 4 R   

so, the net electric field due to this part of thread at point O is

1

ˆ

||

ˆ

E



E i E j

_



Putting the value of E and E||, we get

We get 1 0 0 ˆ ˆ E i j 4 R 4 R      

(ii) Second part:

A circular arc which is making angle

2



at the centre of curvature O

 The electric field due to this part is

E2 = ₂ 0 q sin 2 R 2    =





₂ 0

R

sin

2

R

2

 





= 0 sin 2 R 2   

crackiitjee.in

90

2



 





2 0 sin 4 E 2 R      = 0

2 2

R





In vector from, 2 2 ˆ 2 ˆ E  E cos 45 i E sin45 j or ₂ 0 0

1

_ˆ

1

_ˆ

E

i

j

2 2

R

2

2 2

R

2















_

_









or ₂ 0 0 ˆ ˆ E i j 4 R 4 R      

(iii) Third part:

This consists of a semi-infinite thin horizontal thread.

0 E 4 R     || 0 E 4 R   

The electric field at point O due to this part is

3 ||

ˆ

ˆ

E

 

E i E j



_

= 0 0 ˆ_i ˆ_j 4 R 4 R  _   

The net electric field at point 0 is

1 2 3

E E





E



E .

Putting the values of

E , E and E ,

_{1 2 3} we get

2

E E



= 0 0 ˆ_i ˆ_j 4 R 4 R  _    0 2 E 4 R    

(b) The thread of problem may be divided into three parts:

(i) First part:

Semi-infinite vertical thread. The electric field at point O due to this part is

1 0 0 ˆ ˆ E i j 4 R 4 R      

(ii) Second part:

This part is semicircular thread of radius R. The electric field due to circular arc is

E2 = 0 sin 2 R 2   

In the case of semicircular arc,

 =  2 0 E 2 R     In vector form, 2 0 ˆ E j 2 R   

(iii) Third part:

Semi-infinite vertical thin thread

The electric field due to this part is

3 0 0 ˆ ˆ E i j 4 R 4 R      

 Net electric field due to thread at point O is

1 2 3

E E





E



E

Substituting the values of

E , E and E ,

_{1 2 3} we get,

E 0.



Q. A very long uniformly charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge of the thread per unit length is equal to . Find the flux of the vector E across the circle area.

Since the solid angle subtended by a right circular cone on its vertex is  = 2(1 – cos ), where  is semivertex angle.

Electric flux through art area ds due to a point charge q. Electric flux  is defined as

E.ds

 

= ₂ 0 q dx cos 4 r  = ₂ 0 q ds cos 4 r  

E.ds

 

= ₂ 0 q ds cos 4 r  0 q 4     

where,  is solid angle subtended by area ds on the point charge.

In similar fashion, electric flux passing through a given area due to an elementary charge dq is 0 dq d 4    

where  is solid angle subtended on the elementary charge dq. We consider an element dx at distance x from centre of circle.

The electric charge on the considered element is dq = dx. (treated as point charge).

The solid angle subtended by circle on the considered element is  = 2(1 – cos ), Here, cos  = 2 2 x R  x 2 2

x

2 1

R

x





   

_



_



_



The electric flux passing through the circle due to considered element is

0 dq d 4     = 2 2 0

2

x

1

dx

4

_R

_x















 

_



crackiitjee.in

 Total electric flux crossing through the circle due to large thread is d  



 = 1 2 2 0 0

x

1

dx

2

_R

_x



















_



Here, put x = R tan dx = R sec2d x 1 2 0 x 0 R tan 1 R sec d 2 R sec         _  _     



 =





1 2 0 0 1 sin R sec d 2      



= x 1 x 1 2 0 x 0 x 0

R _{sec d tan sec d} 2        _{     }    





 =



 



x 1 x 1 x 0 x 0 0 R _{tan sec} 2    _   _{  }    





x 1 x 0 0 R tan sec 2           1 2 2 0 ₀ R x R x 2 R R           _ _ = 2 2 0

R

l

R

l

1

2

R

R

















 

_



_

crackiitjee.in

Thread is very long.  R >> 1 2 2 0

R

l

l

R

1 1

2

R

R l







  





 



 

_



_

2 2 R l may be neglected 0

R l

l

₁

2

R

R

 



  

_





_

 



0 R 2     

Q. Two point charges q and –q are separated by the distance 2l. Find the flux of the electric field strength vector across a circle of radius R.

Electric flux through an area due to a number of point charges is given by

n i i i 1 0

1

_q

4

_

 







The solid angle due to circle on the point charge q is

 = 2 (1 – cos )

Since, charges are opposite sides of the circle. So, net flux passing through circle is

 = 1 – 2

Here, 1 1 0 q 4     =





0

q2 1 cos

4









=





0

q 1 cos

2







Similarly, 2 2 0 q 4     =





0

q2 1 cos

4











=





0

q 1 cos

2









1 2      =





0

q

1 cos







cos  = 2 2 l R l

crackiitjee.in

2 2 0

q

₁

l

R

l





  

_



_

 

_



Q. r

unit length. The threads are separated by a distance . Find the

maximum magnitude of the electric field strength in symmetry plane of this system located between the threads.

The electric field due to a long thread (thin rod) is E =

0 2 r

 

The direction of electric field is along increasing r.

Let both threads are placed perpendicular to the plane of paper (along z-axis) at a separation of l.

Horizontal components of electric field balance each other.

 Net electric field at point P is E = 2E1 cos 

But electric field due to one of the thread is: E1 = 0 2 r   0 2 E cos 2 r     

crackiitjee.in

= 0 cos r    = ₂ 0 y r   E = ₂ 2 0 y l _y 4         

For maximum electric field,

dE

0

dy



After solving, we get y =

l

2

max 2 2 0

l

2

E

l

l

4

4

 

 

 









_

= 0l  

Q. An infinitely long cylindrical surfaced of circular cross-section is uniformly charged lengthwise with the surface density  = 0 cos ,

where  is the polar angle of the cylindrical coordinate system whose z

crackiitjee.in

axis coincides with the axis of the given surface. Find the magnitude and direction of the electric field strength vector on the z axis.

Here, since, surface charge density is given. So, the cylinder may be treated as hollow charged cylinder.

The given cylinder behaves as a long bamboo. So, the strip of the cylinder alone the cylinder behaves as long thread.

The top view of cylinder is like a circular ring.

Now we consider a long strip of width dt = Rd If length of cylinder is l(l

).

The area of strip is dA = ldt = lRd

The charge on the strip is dq = dA

=0 coslRd

= 0 lR cos d

The linear charge density of considered strip is,

dq

l

 

= 0

lR cos d

l



 

crackiitjee.in

= ₀R cos d 

The electric field at point O due to considered long strip is dE = 0 2 R   = 0 0 R cos d 2 R     = 0 0 cos d 2      dE  dEcos i dEsin j ˆ ˆ

The component of electric field along x-axis due to whole cylinder is

Ex = 2 0 dEcos     



= 2 2 0 0 0 cos d 2   _{ } 



= 2 0 0 0 1 cos2 d 2 2     _{ }     



=





2 0 0 0 1 cos2 d 4   _{  } 



= 0 0 2  

crackiitjee.in

The component of electric field due to whole cylinder along y-axis is Ey = 2 0 dEsin     



= 2 0 0 0 cos sin d 2       



= 2 0 0 0 sin2 d 0 4       



 Net electric field at point O is E = E i E j_xˆ _yˆ = 0 0 ˆi 2    i.e., E = 0 0 2  

Q. A ball of radius R carries a positive charge whose volume density depends only on a separation r from the ball's centre 0 (1 –

r

R

),

where 0 is a constant. Assuming the permittivities of the ball and the

environment to be equal to unity, find:

(a) the magnitude of the electric field strength as a function of the distance r both inside and outside the ball;

(b) the maximum intensity Emax and the corresponding distance rm.

This problem can easily be solved by using Gauss's law. i.e., in 0 c

q

E.dS







or 0 c

dq

E.dS









Case-I. When r < R in 0 c

q

E.dS







in 0 c

q

EdS cos0







2 in 0 q E 4 r   in 2 0 q E 4 r    …(i)

where, qin = Total electric charge enclosed in a sphere of radius r

Here, qin = r 2 0 4 r dr  



crackiitjee.in

= r 2 0 0 r 1 4 r dr R    _  _   



= r r 3 2 0 0 0 r 4 r dr dr R    _  _ 





 = 3 4 0 r r 4 3 4R    _  _   = 3 4 0 0 4 r r 3 4R    _    _{ } From equation (1), E = in ₂ 0 q 4 r = 3 4 0 2 4 r r 4 r 3 4R        _{ } 2 0 0 r r E 3 4R      _  _    radially outward 0 0

r

3r

E

1

3

4R







 

_



_

 



where r < R Case-II.

When r = R at the surface of sphere

E = 2 0 0 R R 3 4R    _      = ₀R 4 3 12     _{ }   0 0 R E 12     radially outward. Case-III. When r > R In this case, qin = R 2 0 0 r 1 4 r dr R    _  _   



or qin = 3 4 0 R R 4 3 4R    _  _   or qin = 0 3 1 1 4 R 3 4    _  _   or qin = 3 0

1

4

R

12





= 3 0R 3 

From equation (i), we have

E = in ₂ 0 q 4 r = 3 0 2 0

R

3 4

r



 

3 0 2 0

R

E

12 r



 



0 0

r

3r

E

1

3

4R

 



 

_



_

 



where r < R E = 0 0 R 12   when r = R and E = 3 0 2 0

R

12 r





when r > R

(b) From expression of electric field, the electric field outside sphere decreases with increasing r. Hence, electric field will be maximum inside the sphere.

0 0

r

3r

E

1

3

4R

 





_



_

 



 For maximum electric field,

dE

0

dr



or

1

3r

0

2R





2R

r

3

 

At r =

2R

,

3

electric field will be maximum.

0 max 0

2R

3 2R

3

E

1

3

4R 3





 

_

_{ }

_

_

_

_





_



_

_

_







_

_

= 0 0 R 9  

Q. A system consists of a ball of radius R carrying a spherically symmetric charge and the surrounding space filled with a charge of volume density  =

,

r



where  is a constant, r is the distance from the centre of the ball. Find the ball's charge at which the magnitude of the electric field strength vector is independent of r outside the ball. How high is this strength? The permittivities of the ball and the surrounding space are assumed to be equal to unity.

The electric field on the surface of a sphere is

E = ₂

0 q 4 r

And this result is independent of distribution of charge inside the sphere. Now, we take origin at the centre of sphere. From r = 0 to r = R, space is uniformly charged (volume charge density is constant).

From r = R to r , volume charge density is

.

r



 

For convenience, the system may be assumed as combination of a uniformly charged sphere of radius R and a hollow sphere of inner radius R1 = R and outer radius R2

. The resultant electric field at a point is calculated by vector sum of fields due to both spheres.

1 2

E E





E

Calculation of E1: E2 = 0 ₂ 0 q

4 x at point P at a distance x from centre of sphere.

Here, q0 = charge on the sphere. Calculation for E2:

E2 = ₂ 0 q

4 x where q is charge on the hollow sphere from r = R to r = x.

Analysis for q.

We consider a hollow sphere of radius r (R < r < x) and thickness dr. The volume of considered element is

dV = 4r2dr.

 Electric charge on the considered element is dq =dV

=

4 r dr

2

r



_

= 4rdr

The total charge enclosed from r = R to r = x is q =



dq = x R 4



rdr = 2 2 x R 4 2     _ _ = 2



x2 R2



 Net electric field is E = E1 + E2 (both are in the same direction).

 E = 0 _{2 2} 0 0 q q 4 x  4 x or E =





2 2 0 2 2 0 0 2 x R q 4 x 4 x     

crackiitjee.in

(on putting the value of x) or E =





2 2 0 2 0 q 2 x R 4 x     or E =





2 2 0 2 0 q 2 R 2 x 4 x     

According to problem, result should be independent of x (in the problem, r is taken at the place of r).

If q0 – 2R2 = 0 Then E = 0 2 4   which is independent of x. Hence, q0 – 2R2 = 0  q₀ = 2R2 and E = 0 2  

Q. A space is filled up with a charge with volume density 0

3

r e

where ₀ and  are positive constants, r is the distance from the centre of this system. Find the magnitude of the electric field strength vector as a function r. Investigate the obtained expression for the small and large values of r, i.e. at r3 << 1 andr3 >> 1.

The electric field on the surface of spherical distribution of charge, is

E = ₂ 0 q 4 r  

This result is applicable for uniform as well as non-uniform distribution of charge.

This result does not depend upon charge outside the surface. Draw a sphere of radius r.

Now, calculate the net charge enclosed by the sphere.

For this, we consider a hollow concentric sphere of radius r and thickness dr. The volume of considered element is dV = 4r2dr.

The electric charge in the considered element is dq = dV

= ₀er34 r dr 2 = 4₀r e2 r3dr

 Total charge enclosed by the sphere is

q = 3 r 2 r 0 0 4_ r e dr





_r3



0 0 4 q 1 e 3        2 0 q E 4 r   radially outward

crackiitjee.in

= 0 ₂ r3 0 1 e 3 r   _{  }    

Let us discuss the condition of problem: According to problem, 3 r 1   3 r 3 e _{  }1 r E = 3 0 2 0

r

3

r

 

 

= 0 0 r 3   3 r e _ 0 (whenr3 >>> 1) 0 2 0 E 3 r     

Q.1 Two metal balls of the same radius are located in homogeneous poorly

conducting medium with resistivity (). Find the resistance of the medium between the balls provided that the separation between them is much greater than the radius of the ball.

Sol. ∵ Electric resistance is R =∫

Let us consider a hemispherical shell of radius r and thickness dr whose centre is at O.

The electric resistance of considered element is dR = 

∴ The resistance between sphere is

R = ∫  =  l a a

1

r





_











=  0 1 =  0 ( )1 R =  0 ( )1 But l >>> a, ∴ l – a  l ∴ R =  Alternative method:

Ohm’s law in the vector method is: ⃗⃗ =  Where = current density

At first, spheres are charged by opposite nature of charge of magnitude q. The electric field at the surface at the first ball is E =

Here the second ball is infinitely at large distance from first ball.

So, electric field due to second ball on the surface of first ball will be zero. The current density J = _

J =  Electric current I = ∫ ⃗ dS = ∫ = J∫ = j × 4 =  4 =

But the potential difference between ball is = -

= - . / = ∴ Electric resistance R = (from Ohm’s law)

∴ R =

Q.2 A metal ball of radius a is located at a distance l from an infinite ideally

conducting plane. The space around the ball is filled with a homogenous poorly conducting medium with resistivity ( ). In case a ≪ 1 find:

(a) The current density at the conducting plane as a function of distance r from the ball if the potential difference between the ball and the plane is equal to V; (b) The electric resistance of the medium between the ball and the plane. Sol.

Potential difference between the ball is = V =

q = V (1) (a ) The electric field at the point P is due to charge q is

E =

∴ Net electric field at point p is E = 2E0 cos

or E = cos

or E =

from equ. (1)

or E =

according to Ohm’s law, E = J

∴ J = =

(b) Now we consider a hemispherical shell of thickness dx and radius x whose centre is at the centre of the ball.

R = ∫ = ∫  =  l a 0

1

r





_











=  0 1 =  (∵ l > > a)

Q.3 Two long parallel wires are located in poorly conducting medium with

resistivity ( ). The distance between the axes of the wire is equal to l, the cross section radius of each wire equals a. in the case a ≪ l find;

(a) the current density at the point equally removed from the axes of the wire by a distance r if the potential difference between the wires is equal to V.

(b) The electrical resistance of the medium per unit length of the wires.

Sol.

(a) The electric field at a point P at a distance r from both wires is E =

By ohms law E = J

J = =

(b) The electrical field near first wire is E0 =

J = =

I = j S = j 2 a (per unit length) R = (per unit length) R =

=

=

Ans.

Q.4 The gap between plates of a parallel plate capacitor is filled with a glass of

resistivity ( ) = 100 G.m. The capacitance of the capacitor equals C = 4.0 nF. Find the leakage current of the capacitor when a voltage V = 2.0 kV is applied to it.

Sol. The problem is solved in following steps:

Find resistance and capacity of system separately.

Here R =

Also, C =

∴ = Now draw equivalent circuit

In loop AGEFA, = 0 ∴ q0 = CV In loop ABDFA, - IR + V = 0 ∴ V = IR ∴ I = = = =

Here current I through equivalent resistance of the system is known as leakage current.

According to problem, C = 4 ×

 = 100 × m

= dielectric constant for glass = 6 Substituting the values we get,

I = 1.5 × A = 1.5 A

Q.5 Two conductors of arbitrary shape are embedded into an infinite

homogenous poorly conducting medium with resistivity () and permittivity ( ). Find the value of a product RC for this system, where R is the resistance of the medium between the conductor, and C is the mutual capacitance of the wires in the presence of the medium.

Sol. Here Ohm’s law in the vector form is applicable i.e. ⃗⃗ = 

∵ Conductors are oppositely charged by charge q.

The electric field at the surface of conductor is E = in the direction of area.

Here =

But E = J (here ⃗⃗  ) ∴ J = _

Since, electric field and area is in same direction so, angle between area vector and current density vector should be zero.

∴ J = ∴ I = JS (where s is area) Or I = _ or V = IR

(Here V is potential difference between conductors)

Or = IR

∴ RC = = = =

RC = Ans.

Q. 6: A conductor with resistivity bounds on a dielectric with permittivity , at a

certain point A at the conductor’s surface the electric displacement equals D, the vector D being directed away from the conductor and forming an angle a with the normal of the surface. Find the surface density of charges on the conductor at the point A and the current density in the conductor in the vicinity of the same point.

Sol.

∴ Surface charge density is = = D (normal component)

And D = E

∴ E = at point

The tangential component of electric field inside conductor is E0 = E

By Ohm’s law, E0 =

∴ J = = =

Q.7 The gap between the plates of a parallel plate capacitor is filled up with an

in-homogenous poorly conducting medium whose conductivities varies linearly in the direction perpendicular to the plates from 1 = 1.0 pS/m to 2 = 2.0 pS/m.

Each plate has an area S = 230 cm2, and the separation between the plates is d =

2.0 mm. Find the current flowing through the capacitor due to a voltage V = 300 V.

Sol. ∵ R = ∫ and  = _ Here  is linear function of distance x.

 = mx + K when x = 0,  = 1 1 = m × 0 + K K = 1 when x = 0,  = 2 ∴ 2 = md + K 2 = md + 1 ∴ m = ∴  = . / x + 1

Now solve the problem in following ways:

Find equivalent resistance, we consider an element of thickness dx at distance x from first plate.

The electric resistance of considered element is dR =  = _ =

0. / 1

The equivalent resistance between plates is R = ∫ = ∫ 0. / 1 = ( )

Now draw equivalent circuit of system :

The system behaves as parallel combination of a resistor of resistor R and a capacitor of capacitance C.

V = IR

I =

Putting the value of R1, we get

I = ( )

Putting the values, we get I = 5 nA

Q.8. Demonstrate that the law of refraction of direct current lines at the

boundary between two conducting media has the form = where 1

and 2 are the conductivities of the media, and are the angles between the

current lines and the normal of the boundary surface.

Sol. Since the electric charge is neither pilling up nor disappearing at a boundary,

the current toward the boundary on one side must be equal to that away from it on the other side.

By conservation charge, I1 = I2 or ⃗⃗ . d ⃗ = ⃗⃗ . d ⃗ or (J1 cos 1) dS = (J2 cos 2) dS or J1n = I2n ……….. (i) (normal component of J) By Ohm’s Law, J =  E 1 E1n = 2 E2n ……… (ii)

But there is no source of emf. ∴ ∫ ⃗⃗ . d ⃗ = 0

Here E_1n = E_2n

 =

or = ……….. (iii)

(∵ J1t and J2t are horizontal components of J1 and J2 respectively)

From equation (i) and (iii), we get ∵ 1 cot 1 = 2 cot 2

Here, 1 = 1 and 2 = 2

∴ = ∴

=

Q.9. Two cylindrical conductors with equal cross sections and different

resistivities 1 and 2 are put end to end. Find the charge at the boundary of the

conductors if a current I flows from conductor 1 to conductor 2.

Sol. By Ohm’s Law, ⃗⃗ = 

The problem can be solve in following ways: Find electric fields in both conductors. The current density in first conductor is

J1 = (Along the length of conductor)

Similarly, J2 = along the length of conductor.

The electric field inside first is E1 = 1 J =

Similarly, E2 = (Along length of conductor)

Now consider small cylindrical region at boundary The area of cylindrical region is

S =  R2

The net flux through this region is

 = 1 + 2 = ⃗⃗⃗⃗ . ⃗ + ⃗⃗⃗⃗ . ⃗

= E1 S cos 180 + E2 S cos 0

= - E1 S + E2 S = (E2 – E1) S

According to Gauss’s Law,

 = or (E2 – E1) = =

∴ The surface charge density at boundary is

 = (E2 – E1)

Substituting the value E1 and E2, we get

∴  = (2 j - 1j) or  = (2 - 1) j

∴  = (2 - 1)

or R2 = (2 - 1) I

or r = (2 - 1) I

Charge at the boundary q = (2 - 1) I

Q.10. The gap between the plates of a parallel plate capacitor is filled up with

two dielectric layers 1 and 2 with thickness d1 and d2, permitivities 1 and 2, and

1 and 2. A dc voltage V is applied to the capacitor, with electric field directed

from layer 1 to 2. Find , the surface density of extraneous charges at the boundary between the dielectric layers and the condition under which  = 0

Sol. When electric field is uniform, then

E = This problem can be solved in following ways: Find electric field in both dielectric separately

V = V1 + V2 (∵ E1 = and E2 = )

or V = E1 d1 + E2 d2 ……… (i)

If leakage current through capacitor is I, then J1 = and J2 =

According to Ohm’s law,

E1 = 1J1 and E2 = 2J2 J1 = _ and J2 = _ But J1 = J2 = ∴ _ = _ E2 =   E1

From eqn (i), we have E1 d1 + E2 d2 = V

E1 d1 +   E1 d2 = V ∴ E1 =    Similarly, E2 =   

Now apply Gauss’s law at the boundary:

We consider a cylindrical region of cross sectional area dS near the boundary. According to Gauss’s law, ∫ ⃗⃗ . d ⃗ =

or - 1 E1 dS + 2 E2 dS = or (2 E2 - 1 E1) = or (2 E2 - 1 E1) =  ∵  = 0 (2 E2 - 1 E1)

Putting the values of E1 and E2, we get

 = _{ } (22 - 11)

For  = 0, 22 = 11

Q.11. An in-homogenous poorly conducting medium fills up the space between

the plates 1 and 2 of a parallel plate capacitor. It’s permittivity and resistivity vary

from values 1, 1 at plate 1 to plate 2. Find the total extraneous charge in the

given medium.

Here, ρ varies linearly from first plate second plate. ∴ ρ = mx + K When x = 0, ρ = ρ1 ∴ ρ1 = m × 0 + K ∴ K = ρ1 When x = d, ρ = ρ2 ∴ ρ2 = md + ρ1 ∴ m = . / x + є Similarly, є = . / x + є1

The electric field at left surface of considered element is E = ρJ E = 2. / 3 J ∴ E = ρJ ∵ dE = J d ρ ∵ ρ = ( ) x + ∴ = . / ∴ = . / Similarly, є = . /

According to the solution of previous problem, σ =

For considered element, = + d = + d =

∴ dσ = ( )( )-

= + = But may be neglected.

∴ dσ = + or dσ = 2. / 3 J . / 2. / 3 J . / But J = ∴ Sdσ = I 2. / 3 . / 2. / 3 . / or = 2. / 3 . / +2. / 3 . / = . / . / 2. / 3 . / or ∫ = . / ∫ 2. / 3 +. / ∫ 2. / 3 or = . / 2. / 3 + . / 2. / 3 or =. / 2( ) 3 + . / 2( ) 3 = . / ( ) + . / + . / ( ) +. / = - - + + - + - - + + - = - - + + - + - or = - ∴ Q = ( )

Q.12 A long round conductor of cross-sectional area S is made of material whose

resistivity depends only on a distance r from the axis of the conductor as ρ = 

, where  is a

constant, Find:

(a) The resistance per unit length of such a conductor.

(b) The electric field strength in the conductor due to which a current I flows through it.

∵ = ∫

Sol. Let us consider a long co-axial hollow cylinder of radius r and thickness dr

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The cross-sectional area of considered element is dA= 2 .

∴ Electric resistance between ends of conductor of considered element is dR = ρ = ρ = 

The system may be assumed as parallel combination of a number of such types of elementary resistance. ∴ = ∫ = _ ∫ or = _ . / ∴ R =  But S = ∴ R = . /

∴ Resistance per unit length is =  According to Ohm’s law V = IR

But E = =

∴ E = . / I = 

Q.1- A capacitor filled with dielectric of permittivity = 2.1 loses half the charge

acquired during a time interval  = 3.0 min. Assuming the charge to leak only through the dielectric filler, calculate its resistivity.

Sol.

This system behaves as parallel combination of a capacitor C = and a

resistance of R =

∵ At t = 0, charge on capacitor is q0

When t > 0, charge start to leak because to this, charge on plate start to decrease.

Applying loop rule in circuit.

- IR = 0

But I =

(since, charge decrease with

increase of time) ∴ + = 0 or = Or ∫ = ∫ After integrating, q = q0 ⁄

According to problem, when t = , q =

∴ = q0 ⁄ or = ⁄

Or - ln 2 =

∴ = 

After substituting the values, we get = 1.4× 1013 m

Q. 2 a circuit consist of a source of a constant emf  and a resistance R and a capacitor with capacitance C connected in series. The internal resistance of the source is negligible. At a moment t = 0 the capacitance of the capacitor is

abruptly decreased -fold. Find the current flowing through the circuit as a function of time t.

Sol.

This problem is solved by following ways: First draw circuit at t < 0.

In this situation, circuit is in steady state. So, current in the circuit is zero. According to the loop rule,

 - = 0

∵ Q = C 

Now discuss the situation at t = 0. Since, at t = 0, the capacity of capacitor is abruptly decreased.

i.e. C1 = _

Due to this, voltage of the capacitor abruptly increases. This voltage is greater than the emf of cell. So, charge starts to decrease for getting new steady. The direction of flow of charge

Now draw the circuit at an instant t (> 0). At an instant t, the circuit

According to loop rule, – IR -  = 0

Here I = (since charge decreases) ∴  - = R Or C1 – q = RC1 or  – = Or ∫ _ – = ∫ Or [ ( – )]Q = Or 0 _ – – 1 = Or ln _ – _– = Or  – = ( – ) ⁄ Or _ - q = ._ – / ⁄ Or _ - q = . _/ ⁄ Differentiating both sides with respect to t

= ( ) .   / .  / ⁄ Or I =  ( ) ⁄

Q.3- Find the potential difference 1- 2 between two points 1 and 2 of the circuit.

If R1 = 10, R2 = 20, 1 = 5.0 V and 2 = 2.0 V. The internal resistances of the

current sources are negligible.

Sol.

∵ The potential difference between two points = 1- 2 = - {algebraic sum of rise

up and drop up of the voltage} Here, the distribution of the current

According to loops rule,(in loop ABDEA)

-IR1 + 1 – IR2 + 2 = 0

∴ I =  

∴ 1- 2 = - {algebraic sum of rise up and drop up of the voltage throw path (1A

B2)}.

=-{-IR1 + 1} = IR1 - 1

Substituting the value of I, we get

1- 2 =

(  )



Substituting the values, we get,

1- 2 = -4V

Q.4- Two sources of current of the equal emf are connected in series and have

different internal resistance R1 and R2, (R1 > R2). Find the external resistance R at

which at which the potential difference across the terminals of one of the sources (which one in particular?) becomes equal to zero.

Sol. Here the circuit of the problem.

According to the loop rule,

 - IR1 +  - IR2 – IR =0

∴ I = 

Potential difference between the terminals of first cell is

A- B = - { }

Or A- B =  =





According the problem,

A- B = 0 ∴   = 0 Or = 1 Or 2 R1 = or R = ∴ R2 > R1

So R is negative, which is physically not possible. The potential difference across the second cell is

B- C = - { }

Or B- C = 

= 



Q.5- N source of current with different emf are connected. The emf of the

sources are proportional to their internal resistances, i.e.  = R, where  is an assigned constant. The lead wire resistance is negligible. Find

(a) The current in the circuit;

(b) The potential difference between the points A and B dividing the circuit in n and N –n links.

Sol. By loop rule,

(a) ( ) + ( ) + ……….+upto N terms = 0 N - NIR = 0

∴ I =  =  =  =

(b) A- B = (  ) + (  ) + ………+ upto n terms

= n - nIR = nR - nR = 0

Q.6- In the circuit, the source have the emf’s 1 = 1.0V and 2 = 2.5V and the

resistance have the values R1=10 and R2 = 20. The internal resistances of the

source are negligible. Find a potential difference A- B between the plates A and

B of the capacitor C.

Sol.

Since, in steady state, the current through the capacitor branch is zero. Also charge on capacitor is maximum.

The current distribution in the circuit. Form KVL, in closed loop ABEFA,

1 – IR1 –IR2 -2 =0

∴ I =  

G- D = {1 – IR2} (through GABD)

= 2 3 (through capacitor branch)

Or 1 + IR1 = 2 3 ∴ = -IR1=.   / R1 ∴ A- B = 2 3 = = .   / R1

By substituting the values, we get

A- B = -0.5 V

Q.7- In the circuit the emf of the source is equal to = 5.0V and the resistance are equal to R1 = 4.0 and R2 = 20. The internal resistance of the sources equals R =

0.10. Find the current flowing through the resistance R1 and R2. Sol. The distribution of current in the circuit:

By KVL, in closed loop AGEFA,

 – IR –I1R1 = 0

∴ I =  – ………(i)

In loop GBDEG, -(I- I1) R2 + I1R1 = 0 ………(ii)

After substituting the value of I, we get I1 =



= 1.2A

From equation (ii), we get

I –I1 =

Substituting the value of I1, we get

I –I1 =



On substituting the value we get,

I2 = 0.8A

Q.8- Find the current flowing through the resistance R in the circuit. The internal

resistances of the batteries are negligible.

Sol. Here the distribution of the current in the circuit is

In loop (1),

0 – (I –I1) R1 = 0 ………(i)

In loop (2),

-I2 R3 –I1R2 + (I –I1) R1 = 0 ……..(ii)

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In loop (3),

 + I2 R3 – (I1 -I2) R = 0 ………(iii)

By solving equation (i), (ii), (iii), we get I1 –I2 =

( ) 

( ) ……..(iv)

Q.9- Find a potential difference A- B between the plates of a capacitor C in the

circuit shown in . If the sources have the emf’s 1 = 4.0V and 2 = 1.0V and the

resistance are equal to R1= 10, R2 = 20, and R3=30. The internal resistances

of the sources are negligible.

Sol. The current through the capacitor branch is zero in steady state. The

distribution of the current in the circuit. In loop (1), (DEFGKD)

1 –IR1 –I2R2 = 0 …….(i)

In loop (2)(GHJKG)

2 – (I –I1) R3 + I1R2 = 0 ……..(ii)

After solving equation (i) and (ii), we get

I =  ( ) 

Potential difference between the points F and H through paths FGH.

F- H = - {algebraic sum of rise up and down up of voltage}

= - {-IR1+2} = IR1 -2

By substituting the values of I, we get

F- H =  ( )  ( ) Or H- F =  ( )  ( ) Form F = B, H = A ∴ B - A = H- F =  ( )  ( )

By substituting the values, we get

B - A = -1V

Q.10- Find the magnitude and direction of the current flowing through the

resistance R in the circuit shown in . If the emf’s of the sources are equal to  = 1.5V and  = 3.7 V and the resistance are equal to R1 = 10, R2 = 20 and R =

5.0. The internal resistances of the sources are negligible.

Sol. The distribution of current in the circuit

In loop (1), from KVL   ( ) = 0 Or   =0 Or   ( ) =0 ∴ I =  + . / ……(i) In loop (2), ( )  = 0 Or ( )  = 0

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