physics 03 – Electrostatics

1. The wedge-shaped surface in figure is in a region  of uniform electric field E0 along x axis.The net electric flux for the entire closed surface is  

(a)  9 E0 
(b) 15 E0
(c) 12 E0
(d)  zero

Since, field is uniform, the net flux for the closed surface is zero.
∴ (d)

2.A block of mass m carrying a positive charge q is placed on a smooth horizontal table, which ends in a vertical wall situated at a distance d from block. An electric field E is switched on towards right. Assuming elastic collisions, find the time period of resultant oscillation.

(a)  √2qEd/m                                                                    (b) √8md/qE

(C) √2md/qE                                                                    (d) √md/qE

 Acceleration of the block a = qE /m
d = (½ ) at2


Required time = 2t = √8md /qE

∴ (b) 

3.Three uncharged capacitors of  capacities C1 , C2 , C3 are connected as shown in figure to one another and to points A, B and C at potentials V1 , V2 and V3 . Then the potential at O will be 

(a) V 1C 1+V 2C + V 3C 3   / C + C 2 +C 3

(b) V + V2 +V3   / C1+C2+C3  

(c) V 1 ( V2 +V3)  /  C1 (C2+C3  

(d)  V1V2V3 / C1C2C3

 Taking into account the relation between capacitance,
voltage and charge of a capacitor, we can write the following
equations for the three capacitors.

  V1-V0= q1/c1   ,           V2-V0=q2/c2,                   v3-v0=q3/c3

 

where C1, C2 and C3 are the capacitances of corresponding
capacitors and q1, q2 and q3 are charges on the plates.
According to charge conservation law, q1 + q2 + q3 = 0 and
hence the potential V0 of the common point is

V0   = V1C1+V2C2+V3C3C1+C2+C3

∴ (a)

4.Following operations can be performed on a capacitor.

A: connect the capacitor to a battery of emf E.
B: Disconnect the battery.
C: Reconnect the battery with polarity reversed.
D: Insert a dielectric slab into the capacitor.
Now choose the incorrect option (s)

(A) in action ABC (perform A, then B and then C), the stored electric energy remains unchanged and no thermal energy is developed.

(B) the charge appearing on capacitor is greater after the action ADB than after the  action ABD.

(C) the electric energy stored in the capacitor is greater after the action DAB than the action ABD.

(D) the electric field in the capacitor after the action AD is same as that after the action DA. 

Explanation: (A)
Use the concept of capacitor, charging, discharging etc.

5. Gauss’s law is frequently written in the following form:

  →  →

∫ E.dS = qenc / ε0  Where, the symbols have their usual meanings

 

(A) this law is true for all closed surfaces

(B) it is true only in vacuum

(C) it is true only when the charge distribution is symmetric

(D) it is true only when the electric field is symmetric

 (A)
Gauss’s law is true for all closed surface.

                                                                                            →           

6.If Vo be the potential at origin in an electric field E = Exi ˆ+ Eyjˆ  , then the potential at point P(x, y) is

 

(A) V0+xEx +yEy                                                    (B) V0+xEx -yE

(C) V0– xEx – yEy                                                    (D)  √(x2+y2) . √Ex+ E –  V0

EX= -dV/dx and Ey = – dV/dy

taking X -component 

vov a  dV = – ox  EX dX

 

Vp– V0 = -Ex(X) 

Taking y-component

vovB  dV = – oy  Ey dy

VB– V0 = -Ey(y) 

Adding,VB= V0 – xEx yEy

7. An electron of mass me, initially at rest, moves through a certain distance in a uniform electric field in time t1. A proton of mass mp, also initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t2/t1 is equal to

 

A) 1                                                B) (me/ mp )½

C) (mp/ me )½                             D) (mp/ me

C) 

Acceleration in uniform electric field

a= qE/m 

If t is time for a distance d,

d= 1/2 (qE/m) t 

or , t= √2dm/qE

SO,t1/t2 = √me/mp

8. The effective capacitance between A and B is ( each

capacitor is of 1 μF) 

A) 15/2  μF

B) 17/3  μF

C) 13/8  μF

D)19/8  μF

Circuit can be redrawn as

∴ Ceq =13/8 μF

∴(c)

9. Two identical thin rings, each of radius R metres are coaxially placed at a distance R metres apart. If Q1 and Q2 charges are spread uniformly on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

 

A) zero 

B) q(Q1-Q2 ) ( √2 -(1)  ) / √2 ( 4πε0 R )

C) q√2 (Q1+Q2 ) / ( 4πε0 R )

D) q(Q1+Q2 ) ( √2 +(1)  ) /  / √2 ( 4πε0 R )

VB = KQ 2/R   +    KQ1/√2 R ,                       VA = KQ 1/R   +    KQ2/√2 R , 

∴V – VB =  KQ 1/R ( 1 –  1/√2 ) + KQ 2/R (1/√2  – (1) )

∴V – V= (K/R) (1  –  1/√2 ) (Q1– Q) ,Where K =1/4πε

 

∴W= q (V – V

 

∴(b)

10. A and B are two concentric spheres If A is given a charge Q while B is earthed as shown in figure, then

(A) the charge  densities of A and B are same

(B) the field inside and outside A is zero

(C) the field between A and B is not zero

(D) the field inside and outside B is zero

Since, B is grounded, therefore VB = 0

kQA/R   +   kQB/R    = 0   ( R=radius of shell B )

 

QB = -QA   

now take 

QA    =Q 

σ A = – Q/ 4πR2

BUT   σ A  = Q/ 4πR2

 

∴ (A) is not correct 

Apply Gauss’s theorem. Only (C) is correct.

 

11. Electric charges q, q and – 2q are placed at the corners of an equilateral triangle ABC of side L. The magnitude of electric dipole moment of the system is

 

A) qL                                             B) 2qL 

C) (√3) qL                                     D) 4qL

As shown the three charges are equivalent to two dipoles of magnitude q L.

∴ Equivalent dipole moment

= √¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

     (qL )2 +(qL ) + 2  (qL )  (qL ) c0s 60º                  =  √3 qL     

12. If charges q/2 and 2q are placed at the centre of face and at the corner of a cube, then the total flux through the cube will be

 

A) q/ 2ε0                                                                        B) q/ε0

c) q/ 6ε0                                                                         D)q/ 8ε0  

Flux through the cube when q/2 is placed at the centre of face

              q/2              q

  Φ1  =   ____    =        __    

                              

Flux through the cube when 2q is placed at the corner of cube is

 

             2q               q

  Φ2  =   ____    =        __    

                             

 

Now, total flux through the cube

 

Φ= Φ1  + Φ2 = q/  

13. A solid insulating sphere of radius R is given a charge  Q. If at a point inside the sphere the potential is 1.5 times the potential at the surface, this point will be

 

(A) at a distance of 2R/3 from the centre

(B) at the centre

(C) at a distance of 2R/3 from the surface

(D) data insufficient

Vin = (1/ 4πε0  )  Q( 3R2   – r )  /2R3

 

VS =( 1/4πε)  (Q/R) 

given   

            Vin   =  (3/2 )  vs

 

⇒ (3/2 ) ( 1/R)   = 3R2  – r2  2R3

 

OR r = 0

14. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0, . . .,∞  on the x-axis and a charge – q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0, . . .,∞. Here, x0 is a positive quantity. Take the electric potential at a point due to charge Q at a distance r from it to be Q/ 4πε0r Then, the potential at origin due to the above system of charges is

 

A) 0                                                                   B) ∞

C) q/ 8πεx0 In 2                                           D) q In 2 / 4πεx0           

V= q/4πε ( (1/x0 ) – ( 1/2x) +( 1/3x) – (1/4x)+ ……… )   =  q/ 4πεx 0  1- (1/2) + (1/3) –  (1/4 )+ ……. ) =  q/4πεx in (2)

15. Find the charge on an iron particle of mass 2.24 mg, if 0.02% of electrons are removed from it.

 

(A) -0.01996 C                             (B) 0.01996 C

(C) 0.02 C                                     (D) 2.0 C

Explanation: B 

As  mass/ atomic wt .   = no.of atoms/ avogadro no. =24×10 18  ×26 electrons.

n= no.of electrons removed = 24×10 18  ×26 × (0.01/ 100) = 1248 ×10 14 electron

∴Q=ne = (+ve charge ) = 0.01996 c

16 Two small metallic spheres each of mass ‘m’ are suspended together with strings of length ‘l’ and placed together. When a quantum of charge ‘q’ is transferred to each, the strings make an angle of 90º with each other. The value of ‘q’ is 

A) l √————-

        πε0 mg

B) l √————-

        2πε0 mg

C) l √————-

        4πε0 mg

D) l √————-

        8πε0 mg

Explanation: D

T COS 45 º  = FE

 

= Q2   /  4πε ( 2l)  cos  45 ° 

 

= q2 / 16 πε 2 (1/2)    = q2 /8 πε 2  

 

T sin 45°  = mg 

 

l = mg   8  πε 2   /q2

 

∴q = l  √ 8πε0 mg . 

17 .Two connected charges of +q and – q are at a fixed distance AB apart in a non- uniform electric field, whose lines of forceare shown in the figure. The resultant effect on the two charges is

(A) a torque in the plane of the paper and no resultant force.

(B) a resultant force in the plane of the  paper and no torque.

(C) a torque normal to the plane of the paper and no resultant  force.

(D) a torque normal to the plane of the paper and a resultant force in the plane of the paper.

Explanation: B

UP = ( – 1/4πε0 )qq (1/2R ) – ( 1/R) ) = (1/4πε0 ) ( qqo/2R)

 

Change in potential energy =

gain in kinetic energy =

 

((1/4πε0 ) (qqo/2R)

 

18.A ring of radius R carries a charge +q. A test charge -q0 is released on its  axis at a distance √3R from its  centre. How much kinetic energy will  be acquired by the test charge when  it reaches the centre of the ring?

 

A) (1/4πε0  ) (qq0 /R)                                                                 B) (1/4πε0  ) (qq0 /2R )                      

C)  (1/4πε0  ) (qq0 /√3R )                                                           D)  (1/4πε0  ) (qq0 /3R )                                                                                              

Explanation: C

→                 →

E = K q/ r  r  

Electric field due to 10-9   C   at (3, 1, 1) 

E = K ×( 10 -9 / 11 √11 ) ( ( 3iˆ + iˆ + kˆ 

Electric field due to Q at (3, 1, 1)

E2 = K ×( Q/ 3 √3 ) ( ( iˆ + iˆ + kˆ 

 

EX =0 

⇒ ( 3×10 -9  / 11√11  ) =0 

⇒ Q =  -4.3 × 10-10   C

 

19.The figure shows a spherical capacitor with inner sphere earthed. The capacitance of the system is

A) 4πε0 ab / b-a 

B) 4πε0 b2 / b-a 

C) 4 πε0 / b+a 

D)none of the above

Explanation: B

The potential on the outer sphere is V (assume). Thus we can consider two capacitors between the outer sphere and inner sphere C1 and outer sphere and earth C2. These two capacitors are in parallel.

thus C 1   = 4πε( ab / b-a)

       C2 = 4πε b

       C = 4πε( ab / b-a)   + 4πε b

        ⇒ 4πε0  b2  /  b-a

INTEGERS TYPE QUESTIONS

20.. An electron is projected with a  velocity of 1.186× 10 7   m/s, at an angle θ with the x-axis, towards a large metallic plate kept 0.44 mm away from the electron. The plate has  a surface charge density – 2 × 10 -6 C/m2 . Find the minimum value of θ, or which it fails to strike the plate.

i)40° 

ii)60° 

iii)90° 

iv)70° 

Explanation:

60°  

E= σ /ε0     ,   F = eσ /ε0    

By law of conservation of energy,

          (½ ) mv2  = (½ ) mv2  sin  θ + eσx /ε0

 

    ⇒(½ ) mv2  cos θ + eσx /ε0

Putting the values of σ, m, v and ε0  ,    we get 

              cos θ = 0.248 

              cos  θ =0.5 

    and  θ min =60°

 

 

21. A small ball of mass 2 × 10-3  kg having a charge of 1 μC is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution in a vertical circle. [Take g = 10m/s2  ]

 

i)5.76 m/s

ii)5.56m/s

iii)5.12 m/s

iv)5.96 m/s

Explanation: 5.96 m/s

For looping the loop, at the highest point,

T+ mg – ( q2/4πε0l ) =  mv2 / l 

⇒ V2 = 3.5 (m/s)2      ………..(i)

but the electrostatic work done when  particle goes from bottom to top = 0

Hence,  ½ mu  =½ mv2  + 2mgl 

⇒ u = √35.5 m/s = 5.96 m/s

22. In the adjacent circuit, Each capacitor has a capacitance of  μF. Find the charge that will flow through MN when the  switch S is closed.

 

i)333.3 μC

ii)389.3 μC

iii)365.3 μC

iv)300.3 μC

Explanation:

When S is open:

Equivalent capacitance = C = 2C/3 

∴ Q1 = C1 E = (  2c/3   ) E

 

when S is closed ; 

 

Equivalent capacitance = C2 = 2C

 

∴ Q2 = C2 E =2CE

 

 

∴ Charge flowing through MN =| Q1 – Q2 | = | (2C/3 ) E – 2CE |

 

   =  4/3 CE =( 4/3 )  × 5 × 50 μC = 333.3 μC 

 

23.Four charges + q, + q, – q and – q are placed respectively at the corners A, B, C and D of a square of side a = ( √5 – (1) )  cm  arranged in the given order. If E and P are the midpoints of sides BC and CD respectively, what will be the work done in carrying a charge q0 from O to E and from O to P?

( take q = 10μC, q0 = √5 μC ) 

 

i)J =39J 

ii)J =46J 

iii)J =66J 

iv)J =36J 

Explanation:

ABCD is the given square of side a. The charges are placed at the corners as shown. O is the midpoint of square.

OA = OB =OC=OD =r (say) = a/ √2

Potential at O due to the charges at the corners =1/ 4πεq/r   +   q/r   –   q/r   –   q/r  ) = 

Therefore, O is at zero potential. The electric field at O due to charge

at A

   =  (1/ 4πε0 .( q/r 2)  along OC 

To find the work done in carrying a charge e from O to E

Potential at O = 0

Potential at E = q/ 4πε0  [   1/AE    +     1/ BE    –   1/ DE     – 1/ CE    ]

Since, AE = DE and BE = CE the summation in bracket vanishes. So

potential at E = 0.

Hence no work is done in moving the charge from O to E.

To find the work done in carrying the charge from O to P Potential at 

    P= q/ 4πε0  [   1/AP    +     1/ BP    –   1/ DP     – 1/ CP   ]  = 2q/ 4πε0  [   1/AP    –    1/ DP     ] 

Now, AP=  (AD2 + DP2 )  =   ( a2 + (a/2 )2    =( (√5 )  /2  )  × a           

DP = a/2 

Potential at P

V = 2q / 4πε [  2/√5 a    –    2/a   ]    = ( 2q / 4πε0 ) ×  (2 – 2 √5) /√5 a    = ( q / 4πε0 ) .  4(1 – √5)  √5 a

Potential difference between O and P = 0 – ( q / 4πε0 ) .  4(1 – √5)  /  √5 a   = ( q / 4πε0 ) .  4( √5  – 1 )  /  √5 a 

Work done in carrying a charge e from O to P =

( 1 / 4πε0 ) .  4qe( √5  – 1 )  /  √5 a 

J =36J 

24.A particle of positive charge Q = 8q0, is having a fixed position P. Another charged particle of mass m and charge q = 10 μC moves at a constant speed in a circle of radius r1= 2 cm with centre at P. Find the work t hat must be done to increase the radius of circle to r2 = 4 cm.

 

i) 94 J 

ii) 80 J 

iii) 70 J 

iv) 90 J 

Explanation:

Let q orbit round Q in a circle of radius r.

K.E. of orbiting particle =( 1/2  )  mv 2   ‘……… ( i ) 

 

Where, v is orbital velocity.

Potential energy of q = – (1/4πε )  qQ /r  ………..(  ii )

 

P.E. is negative since q is negative.

 

Electrostatic attraction on q = (1/4πε )  qQ /r2   …….. ( iii ) 

 

This is used as centripetal force required for circular motion.

mv2 / r     =  (1/4πε )  qQ /r2

mv2   =  (1/4πε )  qQ /r     ………..( iv )

From (1) and (4)  

 

K.E.  =( 1/2 )(mv2  )   =  (1/4πε )  qQ /2r

 

Total energy of the orbiting charge

 

= K.E. + P.E.= ( 1/2 ) ( (1/4πε )  qQ /r  )  –  (1/4πε )  qQ /r   = -1/2 ( (1/4πε )  qQ /r   

 

The total energy of q when in orbit of radius r

E 1   = -1/2 ( (1/4πε )  qQ /r1   

 

 

When it is in orbit of radius r

  E 2   = -1/2 ( (1/4πε )  qQ /r 2  

 

The work done on q = change in energy

= E 2    –  E   

= -1/2 ( (1/4πε )  qQ /r 2  )   –   (  -1/2 ( (1/4πε )  qQ /r 1  )   = Qq/8πε (  1/ r1  –  1/ r2   

 

= 8q0 2 /8πε0    (  1/ r1  –  1/ r2   )  =  90 J  

25.A ball of mass m = 100 gm with a charge q can rotate in a vertical plane at the end of a string of length l = 1 m in a uniform electrostatic field whose lines of force are directed upwards. What horizontal velocity must be  imparted to the ball in the upper position so that the tension in the string in the lower position of the ball is 15 times the weight of the ball?

(given qE = 3 mg) 

i)50m/s

ii)40m/s

iii)30m/s

iv)20m/s

Explanation:

As per principle of conservation of energy,

 

K.E at B+ P.E .at B = K.E .at A + P.E .at .A

 

Gain in K.E. = K.E. at A –  K.E. at B

= (1/2 ) m ( v2 2 –  v 1  2 ) ……. ( i)

 

Loss in P.E. = P.E. at B – P.E. at A.

 

= loss in gravitational P.E. at B – gain in electrical energy at A 

 

= mg(2 l ) –  (qE). 2l = (mg – qE)2 l   ……. (ii) 

 

P.E. at B – P.E. at A = K.E. at A – K.E. at B

i,e (mg = qE) 2 l = (1/2) m ( v2 2 –  v 1  2 ) …. ( iii) 

 

Centripetal force at A =  mv2  / l   = ( T 2 + qE – mg ) ….. ( iv) 

 

From equation (3)  mv2  = 2 ( mg -qE ) 21 + m v12

From equation (4)  mv2 2   = 1 ( T 2  + qE-mg )

i.e., 2(mg – qE)2 l +mv1   = 1 ( T 2  + qE-mg ) ……. (v) 

 

Given in problem, T2= 15 mg

∴ 4 mg – 4qE + ( m/l  ) v12  = 15 mg + qE  – mg

or (m/l ) v12 = 10 mg +5qE 

or v12  = l/m   ( 10 mg + 5qE )

 

Horizontal velocity to be imparted to the ball,

 

V = [ 1/ m ( 10mg + 5 qE )]1/2  = 50m/s  

26.Two capacitors are first connected in parallel and then in series. If the equivalent capacitances in the two cases are 16 F and 3 F, respectively, then capacitance of each capacitor is

 

(A) 16 F, 3 F                                           (B) 12 F, 4 F

(C) 6 F, 8 F                                             (D) none of the above

Explanation: B

Let C1 and C2 be the individual capacitance

then C1 + C2 = 16 . . . .(i)

c1c   /  c1   +  c2      =   3      ……..(ii) 


From (i) and (ii) C1 = 12 F, C2 = 4F

27.Two dielectrics of equal size are inserted inside a parallel plate capacitor as shown. By what factor does the effective capacitance increase?

 

A) k1k2  / k + k2                                                B) k + k2     /  2

c) 2 k1k2  / k + k2                                                    D) none of the above 

Explanation: B

Here two capacitors are formed, which are in parallel

Area of each = A/2 and thickness = d

Total capacitance 

 

C=( ε0   (A/2)  / d  )  + ( ε0   (A/2)  / d  )  = (ε0  A / d   ) × (  1  +   2   / 2 )

 

∴ Total capacitance increases by the factor  1  +   2   / 2

28. Two charged balls are attached by silk threads of length l to the same point. Their velocity is K/ √x , where K is a constant and x is the distance between the balls, x is very small in comparison to l. Find the rate of leakage of charge in 10–5 C/s.  (take 1/mg = 10  , k= 4 √2

 

i)20 μC/s 

ii)22 μC/s 

iii)26 μC/s 

iv)30 μC/s 

Explanation:


Let T be the tension in each of the silk threads.

T sin θ = F , T cos θ = mg

tan θ  = F / mg   –  ( q2 / 4 πε0 x2   ) . 1/ mg 

Since θ is small, tan θ = sin θ = x/ 2l 

X = (2/F ) mg =  ( 2l /mg ) . (q2 / 4 πε0 x)

= ( 2l /mg)   . q2 / 4 πε0   = (1/ 2 πεmg  ) q2

x= (  l/ 2 πεmg  )1/3  q2/3 …… (i)

( dx /dq  ) =( dx /dt ) / ( dq /dt ) = l/ 2 πεmg  )1/3  (2/3 )q-1/3 .

dq /dt = q1/3   dx/dt 2/3 l / 2 πεmg1/3 ….. (ii)

It is given,  dx/dt  = K / √x  = k   ( l / 2 πεmg ) 1/6   q 1/3     ….. (iii)

From equations (2) and (3), we get, dq /dt   =  k / 2/3  l / 2 πεmg  )1/2     = ( 3k /2  )  ( 2 πεmg  / l ) 1/2 = 2 × 10 -5  

∴ 20 μC/s 

29. Find the electric flux crossing the wire frame ABCD of length l = 1m width b and whose center is at a distance OP = d (=b/2) from an infinite line of charge with linear charge density λ = π× 10–9 c/m. Consider that the plane of frame is perpendicular to line OP.

i)90

ii)80

iii)30

iv)20

Explanation:

Flux through the element of width dx as shown in

figure is

        →  →

φ = ∫ E. ds , E = λ / (2πε0 d sec θ)  and ds = ldx

Φ = -b/2b/2   λ  cos θ/ (2πε0 d sec θ) ldx 

= λl / 2πε0-b/2b/2   (  d/ √( xd  dx

= λl / πε0  tan-1  ( b/2d ) =90

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