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27 Jan 2009 | 9 min. (1804 words)

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Electrostatics Questions for IITJEE 2009

Questions having one mark each:

  1. Write the S.I unit of
    i. electric field intensity
    ii. Electric dipole moment.
  2. How does the coulomb force between two point charges depend upon the dielectric constant of the intervening medium?
  3. Draw an equipotential surface for a system, consisting of two charges Q, -Q separated by a distance ‘r’ apart.
  4. Show graphically the variation of charge ‘q’ with time’t’ when a condenser is charged.
  5.   What orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium?

Questions having 2 marks each:

  1. Two point charges, q1 =10×10-8 C and q2 = -2×10-8 C are separated by a distance of 60     cm in air.
    i. Find at what distance from the 1st charge, q1, would the electric potential be zero.
    ii. Also calculate the electrostatic potential energy of the system.
  2. Two capacitors of capacitance 6µF and 12µF are connected in series with a battery. The voltage across the 6µF capacitor is 2V. Compute the total battery voltage.
  3. An electric dipole is free to move in a uniform electric field. Explain its motion when it is placed
    i. parallel to the field,
    ii. perpendicular to the field.
  4. An electric dipole of length 10 cm having charges ± 6×10-3 C, placed at 300 with respect to a uniform electric field experiences a torque of 6√3 N-m. Calculate
    i. magnitude of electric field
    ii. the potential energy of the dipole.
  5. Electric charge is distributed uniformly on the surface of a spherical rubber balloon. Show how the value of electric intensity and potential vary
    i. on the surface
    ii. inside and
    iii. outside?

Questions having 3 mark each:

  1. State Gauss’s theorem in electrostatics. Apply this theorem to derive an expression for electric field intensity at a point near an infinitely long straight charged wire.
  2. Explain the underlying principle of working of a parallel plate capacitor.
    If two similar plates, each of area A having surface charge densities +σ and –σ are separated by a distance‘d’ in air, write the expressions for
    i. The electric field at points between the two plates.
    ii. The potential difference between the plates.
    iii. The capacitance of the capacitor so formed.
  3. A 20 µF capacitor is charged by a 30 V d.c. supply and then connected across an uncharged 50 µF capacitor. Calculate
    i. the final potential difference across the combination.
    ii. Initial and final energies. How will you account for the difference in energies?
  4. The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be the effect on its
    i. capacity
    ii. charge
    iii. potential difference
    iv. electric field
    v. energy stored?
  5. Define the term electric potential due to a point charge. Calculate the electric potential at the centre of a square of side √2 m, having charges 100 µC, -50µC, and 20µC and-60µC at the four corners of a square.

Questions having 5 marks each:

  1. Explain the principle on which Van-de-Graaff generator operates. Draw a labeled schematic sketch and write briefly its working. A Van-de- Graaff type generator is capable of building up potential difference of 15×106 V. The dielectric strength of the gas surrounding the electrode is 5×107 Vm-1. What is the minimum radius of the spherical shell required?
  2. An electric dipole is held in a uniform electric field.
    i. Show that no translatory force acts on it.
    ii. Derive an expression for the torque acting on it.
    iii. The dipole is aligned parallel to the field. Calculate the work done in rotating it through 1800 .






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Questions: 1. An electrostatic charge of 21 u C is placed at a distance in air of 15 cm



From a second charge.  The force of attraction between the two charges is 26 N.



A.      Calculate the magnitude of the second charge.



Sol) q1 = 21 u C = 21 x 10-6 C


 


q2 = ?, F = 26N, R = 15cm = 0.15m


 


Using F = q1 q2 / 4πεoR2


 


1/4πεo = 9 x 109 N-m2-C-2


 


Thus F = 9 x 109 q1 q2 / ( 0.15)2


 


or 26 = 9 x 109 q1 q2 / ( 0.15)2




or 26 = 9 x 109 x 21 x 10-6  q2 / ( 0.15)2 = 1.89 x 105 q2 / ( 0.15)2


 


or q2 = 26 x ( 0.15)2 / 1.89 x 105


 


or q2 = 3.1 x 10-6 C = 3.1 μC




B       If the distance between the charges is decreased to 5.0 cm, calculate the magnitude of the new force acting on each ball.


 


Sol) Force between the balls when distance is 15cm = F1 = 26N, R1 = 15cm


 


Let force between the balls when distance is 5cm be = F2 , R2 = 5cm


 


Using relation, F = q1 q2 / 4πεoR2


 


we have F1 = q1 q2 / 4πεoR1 2


 


and F2 = q1 q2 / 4πεoR2 2


 


Thus F1 / F2 = R2 2 / R1 2


 


or 26 / F2 = 52 / 152 = 25 / 225 = 1/9


 


or F2 = 26*9 = 234 N


 


 


Now




2. An electron, moving through an electric field, experiences an acceleration of 6.3 x 103 m/s2.




A. How much electrostatic force is acting on the electron?


Sol) Acceleration a = 6.3 x 103 m/s2.


 


Charge of electron = e = 1.6 x 10-19 C


 


Mass of electron = 9.1 x 10-31 Kg


 


Let Electric force be = F


 


Now F = ma


 


or F = 9.1 x 10-31 x  6.3 x 103 = 5.73 x 10-27 N





B. What is the strength of the electric field?


 


Sol) Let strength of electric field be E


 


Now F = e E


 


but F = 5.73 x 10-27 N  (as calculated above in part A)


 


or E = F/e = 5.73 x 10-27 / 1.6 x 10-19


 


or E = 3.58 x 10-8 N/C








4.      A 3.75 x 10- 9 F capacitor carries a charge of 1.75x10 –8 C




       A.      What is the potential difference across the plates?


Sol) C = 3.75 x 10- 9 F


Q = 1.75x10 –8 C


Now Using Q = CV


we have V = Q/C = 1.75x10 –8 / 3.75 x 10- 9 = 4.67 V










B.       If the plates are 6.50 x 10-4 m a part, what is the strength of the



 Electric field between them?


Sol) Distance between the plates, d = 6.50 x 10-4 m


V = 4.67 V (as calculated above in part A)


Thus E = V/d = 4.67 / 6.50 x 10-4 = 7.18 x 103 V/m







5.      A charge of –3.00 u C is fixed at the centre of a compass.  Two



Additional charges are fixed on the circle of the compass (radius = 0.100m).  The charges on the circle are –4.00 u C at the position due north and + 5.00 u C at a position due east.  What is the magnitude and direction of the net electrostatic force acting on the charge at the centre?


Sol) q = –3.00 u C = -3 x 10-6 C










Bsdsd



           











A



       C(q)




Here BC = AC = 0.1m


Here charge q is placed at C (shown in above figure)


Let at point B which is north to C charge put be = q1 = -4 uC = -4 x 10-6 C


also at point A which is east to C charge kept = q2 = 5 x 10-6 C


Forces at charge q placed at point C due to charges at B and A are mutually perpendicular to each other.


FB = 9 x 109 q.q1 / R2


or FB = 9 x 109 x (-3 x 10-6) x (-4 x 10-6) / (0.1)2


or FB = 10.8 N long BC


Similarly force on q due to q2 at B = FA = 9 x 109 q.q2 / R2


or FA = 9 x 109 x (-3 x 10-6) x (5 x 10-6) / (0.1)2


or FA = 13.5 N along CA


Thus total force F on charge q placed at C is given by


F =


or F =


or F = 17.28 N


The direction of the force is such that it makes an angle θ with CA given by


θ = tan-1 FA/FB = tan-1 13.5/10.8 = tan-1 1.25


or θ = 51.34 degrees.


Hence the direction of the force F = 17.28 N is such that it makes an angle 51.34 deg with CA




6. A particle (mass = 2.5 x 10-15 kg) with a negative charge of 3.5 x 10-5 C is moved from the positive plate of a parallel plate capacitor to the negative plate, a distance of 2.5 cm.



A. If the potential difference between the two plates is 15 volts, how much work is done in moving the particle?


Sol) m = 2.5 x 10-15 kg


q = -3.5 x 10-5 C


Distance between the capacitor plates, d = 2.5 cm = 0.025 m


Potential difference between the plates, V = 15 volts


Thus intensity of electric field E = V/d = 15/0.025 = 600 V/m


Work done in moving the particle = W = F.d  = qE.d


or W = qE.d = 3.5 x 10-5 x 600 x 0.025 = 5.25 x 10-4 J is the amount of work done in moving the particle.




B. If the particle is released from the negative plate, it will accelerate toward the positive plate.  What will be the velocity of the particle just before it strikes the positive plate?


Sol) m = 2.5 x 10-15 kg


q = -3.5 x 10-5 C


V = 15 volts


E = 600V/m (as calculated above)


Now acceleration of the charged particle = a = qE/m


or a = 3.5 x 10-5 x 600 / 2.5 x 10-15


or a = 8.4 x 1012 m/s2


Using the relation v2 – u2 = 2aS


here S = d = 0.025 m


u = 0


Thus v2 – 0 = 2 x 8.4 x 1012 x 0.025


or v2 = 4.2 x 1011 m/s


or v = 6.48 x 105 m/s



C. What is the strength of the electric field between the plates?


Sol) Electric field between the plates E = V/d = 15/0.025


or E = 600 V/m




D. What force will be applied to the particle as it accelerates toward the Negative plate?




Sol) Force on the particle F is given by


F = qE


Here, q = -3.5 x 10-5 C


E = 600 V/m


Thus F = 3.5 x 10-5 x 600 N = 2.1 x 10-2 N = 0.021 N

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