chapter 2. electrostatic potential and … 12th/class...3. two charges 2µµµµc and -2µµµc are...

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1. Two charges 5×€×€×€×€10 C and -3 ×××× 10 C are located 16 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

• Solution: Let the potential be zero at a point P whose distance is x metre from the charge The charge q1 = 5 × 10-8 C. The charge q2 = -3 × 10-8 C The distance of P from q1 = 5 x 10-8 C is x metre. The distance of P from q2 = -3 x 10-8 C is (0.16 - x) metre.

... Potential at P due to q1 =

... Potential at P due to q2 = But the total potential at P should be zero.

... =

0.15 - x = x

0.15 = 2x, ... x = 0.75m.

2. A regular hexagon of side 10cm has a charge 5µµµµC at each of its vertices. Calculate the

potential at the centre of the hexagon?

• Solution: Each side of a regular hexagon = 0.10m Distance of each charge from the centre = Side of the hexagon = r = 0.10 m

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Charge at each vertices 'q' = 5µC = 5 x 10-6 C ... Potential at the centre O of the hexagon

= 6 x

= V = 2.7 x 106 V.

3. Two charges 2µµµµC and -2µµµµC are placed at points A and B 6 cm apart.

(a) Identify an equipotential surface of the system. ~ (b) What is the direction of electric field at every point on this surface?

• Solution:

(a) The equipotential surface will be a plane normal to the line AB joining the two charges. This equipotential surface will pass through the mid-point O of line AB. The potential at any point on this surface is zero. (b) The electric field is diverted from positive to negative charge. It is perpendicular to the equipotential surface.

4. A spherical conductor of radius 12cm has a charge of 1.6 ×××× 10-7 C distributed uniformly

on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18cm from the centre of the sphere?

• Solution: (a) Since the charge of a conductor resides on the outer surface of a conductor the electric field inside the conductor is zero.

(b) The electric field is given as E = where the charge 'q' = 1.6 ×€10-7 C The radius 'R' = 12cm = 0.12m

... E = = 1000 x 102 NC-1

= 105 NC-1 (c) Distance from the centre of the sphere 'r' = 0.18m

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... E = = 444.4 x 102 NC-1 = 4.44 × 104 NC-1.

5. A parallel plate capacitor with air between the plates has a capacitance of 8pf

(1pF=1012 F) .What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6?

• Solution:

The capacitance of the parallel plate capacitor 'C' = Let the distance 'd' between the plates is reduced by half ...the new distance between the plates D = d/2 The dielectric constant = 6

We know that C' = = 2 x 6 x 8pF = 96pF.

6. Three capacitor each of capacitance 9pF are connected in series.

(a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120vsupply?

• Solution:

(a) The total capacitance of the combination in series is given by

= = 3 / 9 Cs = 3pF (b) The potential difference across each capacitor is given by

V = where 'Q' = 120V

... V = = 360 / 9 = 40V.

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7. Three capacitor of capacitances 2pF, 3pF and 4pF are connected in parallel

(a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply?

• Solution:

(a) The total capacitance of the combination in parallel is given by Cp = C1 + C2 + C3 = 2pF + 3pF + 4pF = 9pF (b) The charge on each capacitor is given by Q1 = C1 V Where V = 100 V ... Q1 = 2 × 10-12 ×€100 = 2 × 10-10 V ... Q2 = 3 × 10-12 × 100 = 3 × 10-10 V ... Q3 = 4 × 10-12 × 100 = 4 × 10-10 V.

8. In a parallel plate capacitor with air between the plates each plate has an area of

6 ×××× 10-3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitoris connected to a 100 V supply, what is the charge on each plate of the capacitor?

• Solution:

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Area of the plate 'A' = 6 x 10-2 m2 Distance between the plates 'd' = 3mm = 0.003m ... The capacitance of the capacitor is

C =

= = 1.8 x 10-11 = 18pF. Let the capacitor be connected to a 100 V supply. ... The charge on each plate of the capacitor Q = CV = 18 ×10-12 × 100 = 18 × 10-10 C.

9. Explain what would happen if in the capacitor given in the previous exercise a 3 mm

thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected.

(b) after the supply was disconnected.

• Solution: Capacitance of the capacitor C = 18 pF Dielectric constant of mica sheet K = 6 (a) The capacitance of the capacitor will become K times when the voltage supply remains connected. Hence C' = K ×€C = 6 × 18 pF = 108 pF The charge on the capacitor, q = C' ×€V = 108 ×€10-12 ×€100 = 1.08 ×€10-8 C (b) When the voltage supply is disconnected, the potential difference will decrease on introducing the mica sheet of dielectric constant 6. Hence V' = 100/6 = 16.67 V The charge on the capacitor, q = C' ×€V' = 1.08 ×€10-8 × 16.67 = 1.8 × 10-9 C.

10. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is

stored in a capacitor?

• Solution: Direct Method U = 1/2 CV2

= 1/2 x 12 x 10-12 x (50)2 = 15000 x 10-12

= 1.5 x 10-8 J The capacitance of the capacitor 'C' = 12pF The potential difference 'V' = 50V ... The charge 'Q' in the capacitor is given by Q = C ×€V = 12pF × 50 = 6 × 10-10 C

... The electrostatic energy stored in the capacitor is U = = = 1.5 × 10-8 J.

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11. A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply. And is connected to another uncharged 600pF capacitor. How much electrostatic energy is lost in the process?

• Solution: Capacitance C = 600 ×€10-12 F Supplied voltage V = 200 V Therefore charge q = VC = 200 × 600 × 10-12 = 120000 × 10-12 = 12 ×€10-8 C

Electrostatic energy stored E1 = = × 600 × 10-12 × 200 × 200 = 12 × 10-6 Joules When another capacitor is connected in parallel, then equivalent capacitance

C1 = 2C = 2 × 600 pF = 1200 × 10-12 F = 12 × 10-10 F Charge q = 12 × 10-8 C

Therefore energy stored by the system E2 = = 6 × 10-6 Joules Therefore the electrostatic energy lost in this process is E1 - E2 = 12 ×€10-6 - 6 × 10-6 = 6 × 10-

6 Joules. Direct Method Energy loss = E1 - E2

=

=

= 6 × 10-6J.

12. A charge of 8mc is located at the origin. Calculate the work done in taking a small

charge of -2 ×××× 10-9 C from a point P (0, 0, 3cm) to a point Q (0, 4cm, 0) via a point R (0, 6cm, 9cm)?

• Solution: Initial distance r1 = 3cm = 0.03m Final distance r2 = 4cm = 0.04m q1 = 8mc = 8 x 10-3c, q2 = -2 x 10-9c

Initial potential energy = = J

= - = - 4.8J

Final potential energy =

= = - 3.6J

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... Work done = Final energy - Initial energy = -3.6 - (-4.8)J = -3.6 + 4.8J = 1.2J. The point R need not be taken into consideration because an electric field is conservative. The path is irrelevant.

13. A cube of side B has a charge q at each of its vertices. Determine the potential and

electric field due to this charge away at the centre of the cube

• Solution:

DB2 = b2 + b2 = 2b2

DE2 = DB2 + BE2

or DE =

... DE =

Now DO = DE

=

So, the distance of each vertex from the midpoint of the cube is . In other words, distance of

each charge from the centre O of the cube is ... Potential at O due to charges at the vertices of the cube is,

The electric field at O due to charges at opposite corners such as A and H, B and G, C and F or D and E are equal in magnitude and opposite in direction. So, the net electric field intensity at O is zero.

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14. Two tiny spheres carrying charges 1.5µµµµC and 2.5µµµµC are located 30cm apart. Find the potential and electric field:- (a) at the midpoint of the line joining the two charges, and

(b) at a point 10cm from this midpoint in a plane normal to the line and passing through the midpoint.

• Solution: (a) (i) The potential at the midpoint of the line joining the two charges is given by

= 2.4 × 105 Volt (ii) The electric field at the midpoint O is given by

=

= Vm-1

This electric field is directed towards the charge of 1.5 µC

(b) (i) Potential at P

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whose PA = PB = m = 0.18m

... Potential at P = = 2.0 x 105 Volt. (ii) If E1 is the electric field intensity at P due to 1.5µc then

E1 = 9 × 109 × Vm-1 = 0.42 × 106 Vm-1 If E2 is the electric field intensity at P due to charge of 2.5µc, then

E2 = 9 × 10

= 0.69 ×€106 Vm-1

In ∆ AOP, cos

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... θ = 112.5° cos θ€= cos 112.5° = -0.38.

If is the resultant field intensity, then

= 0.658 × 106 Vm-1 = 6.58 × 105 Vm-1.

15. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

b) Is the electric field inside a cavity (with no charge) zero even if the shell is not spherical, but has any irregular shape? Explain.

• Solution:

(a) The surface charge density on the inner surface of the shall is σ€=

The surface charge density on the outer surface of the shell is σ = (b) By Gauss Law, the net charge on the inner surface enclosing the cavity (without any charge) must be zero. For cavity of arbitrary shape, this is not enough to claim that electric field inside must be zero. The cavity may have positive and negative charges with total charge zero. To dispose of this possibility, we consider a closed loop, part of which is inside the cavity along a field line and the rest inside the conductor. Since field inside the conductor is zero, this gives a network done by the field in carrying a test charge over a closed loop. But this is impossible for an electrostatic field, hence there are no field lines inside the cavity and no charge on the inner surface of the conductor whatever its shape.

16. Show that the normal component of electrostatic field has a discontinuity from one

side of a charged surface to another given by (E2 - E1)n = where n is a unit vector normal to the surface at a point and s is the surface charge density at that point. The direction of n is from side 1 to side 2. Hence show that just outside a conductor, the electric

field is .

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

• Solution: (a) Consider a charged conductor (a cylinder) that has a Gaussian surface about any point P on its surface. The surface charge density at that point is s and has a unit vector n normal to the surface in the outward direction. The cylinder is partly inside and partly outside the surface of the conductor. Let us consider a small area of cross-section dS. Just inside the surface, the electrostatic field is zero, just outside the field is normal to the surface. Thus the contribution to the total flux through the cylinder comes only from the outside cross-section of the cylinder. Hence the electric field inside the surface E1 = 0

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Electric field outside the surface E2 = n

∴∴∴∴(E2 - E1)n = n, where n - is a unit vector. (b) If we consider a closed loop conductor then the electrostatic field inside the conductor is zero, hence there will be no tangential component on the surface, no work is done on moving a small test charge within the conductor and on its surface. If it is not a closed loop conductor, then the electrostatic field E is tangential to the surface and is not normal to the surface. Free charges on the surface of the conductor would experience force and move. Hence the electrostatic field is continuous from one side of the charges surface to another.

17. A long charged cylinder of linear charge density λλλλ is surrounded by a hollow co-axial

conducting cylinder. What is the electric field in the space between the two cylinders?

• Solution: Consider a Gaussian cylinder between the two cylinders of radius r . From symmetry, the electric field E at every point on the surface is directed radially. If q is positive then the direction of E is radially outward. Further, it has the same magnitude at every point of the curved part of the surface because all these points are equidistant from the line of charge. Let the charge enclosed inside the Gaussian surface be q Applying Gauss’s theorem

or

and are contributions to the electric flux by the end caps of the Gaussian surface which vanish as E and dS are perpendicular to each other as these caps.

is the contribution of the curved surface of the cylinder towards the electric flux. So

or

or

rl) =πor E(2

or E = The electric field in the space between the two cylinders are

E = where r is the distance of the point from the common axis of the cylinders. The field is radial, perpendicular to the axis.

18. In a hydrogen atom, the electron and proton are bound at a distance of about

0.53A°. (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from protons.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbits is half the magnitude of potential energy obtained in (a)?

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(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06A° separation?

• Solution: Charge of an electron q1 = -1.6 × 10-19 C q2 = +1.6 × 10-19C Distance between the electron and proton = 0.53 × 10-10 m (a) If Zero of potential energy is taken at infinite separation, then

Potential energy =

=

= (where 1eV = 1.6 × 10-19 J) = -27.2 1eV (b) Kinetic energy = 1/2 × 27.2eV = 13.6 eV ... Total Energy = Kinetic energy + Potential energy = 13.6 eV + (-27.2eV) = -13.6eV Work required to free the electron = -(-13.6 eV) = 13.6 eV. (c) Potential energy at a separation of 1.06 A°

=

= = -13.58 eV = 13.6 eV Potential energy of the system when zero of potential energy is taken at 1.06A°. = -27.2eV - (13.6eV) = -13.6eV ... Required work = -(-13.6) eV = 13.6 eV.

19. If one of the two electrons of the H2 molecule is removed, we get a hydrogen

molecule ion H2 +, the two protons are separated by roughly 1.5A°, and the electron is roughly 1A° from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

• Solution:

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The charge of an electron q1 = 1.6 ×€10-19 C a proton ... q2 = 1.6 × 10-19 C Separation between electrons r12 = 1A° = 10-10 m Separation between the two protons r23 = 1.5A° another proton and = 1.54 × 10-10 m Separation between the two electrons r31 = 1A° = 10-10m If the zero of potential energy is taken at infinity, then

Potential energy U =

= 9 × 10

= J

= (1eV = 1.6 × 10-19J) = -19.2eV The zero of potential energy is taken to be at infinity.

20. Two charged conducting spheres of radii a and b are connected to each other by a

wire. What is the ratio of electric fields at the surface of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions?

• Solution: The ratio of electric field of the first to the second is (b/a). A flat portion may be equated to a spherical surface of large radius, and a pointed portion to one of small radius.

21. Two charges -q and +q are located at points (0,0,-a) and (0,0,a) respectively,

(a) What is the electrostatic potential at the points (0,0,z) and (x,y,0)?

(b) Obtain the dependence of potential on the distance r of the point from the origin when r/a >>1.

(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x axis? Does the answer change if the path of the test charge between the same points are not along the x axis?

• Solution:

Potential at the point = (a) On the axis of the dipole, potential is (± 1/4π ε0)€ρ/(x2-a2) where ρ = 2qa is the dipole moment, the + sign when the point is closer to q and the - sign when it is closer to -q. Normal to the axis, at points (x,y,0) potential is Zero.

(b) The dependence on r is 1/r2. Since r/a >1, ie., r > a ∴q (p) = (c) No work is done in moving a small test charge between the given points along the x-axis. The answer does not change even if the path of the test charge between the same points is not along the

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x axis, because two points is independent of the path connecting the two points. (or)The work done is independent of the path taken.

22. Figure below shows a charge array known as an 'electric quadrupole'. For a point on

the axis of the quadrupole, obtain the dependence of potential on r for r/a>>1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e. a single charge)

• Solution:

Potential at P is given by

V =

=

=

= If r >> a, dropping a2 compared to r2, we get

V = where Q = 2qa2 is called the electric quadrupole moment.

Thus, the potential varies as for a single charge (monopole), as for a dipole, and as for a quadrupole.

23. An electrical technician requires a capacitance of 2µµµµ F in a circuit across a potential

difference of 1kv. A large number of 1µµµµ F capacitor are available to him each of which can withstand a potential difference of not more than 400 V . Suggest a possible arrangement that requires the minimum number of capacitors ?

• Solution: '18' 1µF capacitors can be arranged in 6 parallel rows, each row consisting of 3 capacitors in series.

24. What is the area of the plates of a 2F parallel plate capacitor, given that the

separation between the plates is 0.5 cm? [You will realize from your answer why ordinary

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capacitors are in the range µµµµ F or less. However, electrolytic capacitors do havlarger capacitance (0.1F) because of very minute separation between the conductors.]

• Solution:

The capacitance C = C = 2F Area of the plate A =? Distance of separation between the plates d = 5 × 10-3 m Permitivitty ε0 = 8.854 × 10-12 N-1 m-2 C2

Therefore A = = 113 × 107 m2 = 1130 km2 (approximately).

25. Obtain the equivalent capacitance of the following network. For a 300 V supply,

determine the charge and voltage across each capacitor.

• Solution: Equivalent capacitance = 200/3 pF The charge Q1 across the capacitor C1 = 10-8 C Voltage across C1 = 100V Charge Q2 across the capacitor C2 = Charge Q3 across the capacitors C3 ... Q2 = Q3 =10-8 C and V2 = V3 = 50V Charge Q4 across the capacitor C4 = 2.55 x 10-8 C And the voltage V4 = 200 V.

26. The plates of a parallel plate capacitor have an area of 90cm2 each and are separated

by 2.5mm.The capacitor is charged by connecting it to a 400 V supply (a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

• Solution: (a) Area of the parallel plate capacitor A = 90cm2 = 90 × 10-4 m2

Distance of separation between the plates d = 2.5 × 10-3 m

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Therefore capacitance C = = 31.9 × 10-12 F

Energy stored = = 2.6 ×€10-6 Joules. (b) Energy per unit volume u = total energy / Volume = 2.6 × 10-6 / (90 × 10-4 x 2.5 × 10-3) = 0.115 J m-3

The capacitance of parallel plate capacitor is given by C = where A is the area of each plate and d is the distance between them. Electric field between the plate of the capacitor when it is charged is

E = Q = Eε0A Therefore the energy stored in the capacitor is

U =

= d

= (Ad) 'Ad' is the volume between the plate of the capacitor. Therefore the energy density, i.e, energy per unit volume, is given by

u =

u = .

27. A 4µµµµ F capacitor is charged by a 200 V supply. It is then disconnected from the

supply, and is connected to another uncharged 2µµµµ F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation.

• Solution:

The original charge Q0 is shared by the two capacitors. Therefore Q0 = Q1 + Q2. Using the relation Q = CV, we have C1V0 = C1V + C2V

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or V = --------(1)

= = 133.33 V Since the second capacitor is uncharged, it has no energy. Therefore, the total energy of the system before connection is E0 = energy of the capacitor

C1 = = = 8 × 10-2 Joules --------(2) The total energy after connection is

E =

= = 5.3 × 10-2 joules

Energy loss = E0 - E = 8 × 10-2 - 5.3 × 10-2 = 2.7 × 10-2 joules.

28. A spherical capacitor consists of two concentric spherical conductors, held in position

by suitable insulating supports. show that the capacitance of a spherical is given by C =

where r1 and r2 are the radii of outer and inner spheres respectively.

• Solution: The figure shows a concentric spherical capacitor with radii r1 and r2.

The outer spherical conductor is connected to earth. When charge +Q is given to the inner conductor, it induces -Q on the inner surface of the outer conductor and +Q on its outer surface. As the outer conductor is earthed, the charge +Q flows to earth. If V is the potential difference between

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the spherical conductors A and B, then capacitance of the spherical capacitor is given by C = .In the space between the two spherical conductors, the electric field is not uniform. It varies with

distance and is given by E = dV = E ×€dr. The potential difference V between the two

conductors a be found by integrating both sides V = -----(1) To find the value of V, we consider electric field E at any point P between the two spherical conductors. Let r be the distance of the point P from the centre O of the conductors. We consider the spherical surface of radius r through point P as the gaussian surface. The gaussian surface encloses the inner conductor and hence the charge enclosed by the gaussian surface is Q. According to Gauss's theorem, the total electric flux through the gaussian surface,

φ = -----(2) As the magnitude of the electric field at every point on the gaussian surface is same, we have φ€= E ×€surface area of gaussian surface φ = E ×€4πr2 -----(3) From equations (2) and (3), we get

E€× 4πr2 =

or E = Substituting for E in equation (1), we get

V =

=

=

= Therefore the capacitance of the spherical capacitor is given by

C = =

= 4πε0 . .

29. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of

radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µµµµ C. The space between the concentric spheres is filled with a liquid of dielectric constant 32. (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere?

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(c) Compare the capacitance of this capacitor with 12cm. Explain why the latter is much smaller.

• Solution: Radius of the outer sphere ro = 13cm = 13 × 10-2 m Radius of the inner sphere ri = 12cm = 12 × 10-2 m Charge of inner sphere q = 2.5 × 10-6 C Dielectric constant εr = 32 The charge +q induces a charge -q on the outer sphere. The potential Vi of the inner sphere = potential due to +q on inner sphere + potential due to -q on outer sphere

or Vi = -----(1) Since the outer sphere is earthed, its potential Vo = 0. Therefore potential difference is

V = Vi - Vo = -----(2)

Therefore, capacitance C = -----(3) (a) Substituting the values of ri, ro, εr and ε0 in (3) and solving we get

C = = 5.5 × 10-9 F (b) The potential of the inner sphere

Va =

= = 4.5 × 102 V (c) The potential of an isolated sphere of radius ri is

Vs = -----(4)

∴ Cs = = 4π ε 0ri -----(5) Comparing (4) with (2) we notice that Vs > V. The reason is that the outer sphere has been effectively removed to infinity (ro → ∞ ); hence the potential due to induced charge is absent. Since Vs > V, Cs < C. Substituting for ε0 and ri in (5) gives Cs = 1.3 × 10-11 F.

30. (a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to

each other is the magnitude of electrostatic force between them exactly given by Q1Q2/4 πεπεπεπε0r

2 where r is the distance between their centres? (b) If Coulomb's Law involved 1/r3 dependence (instead of 1/r2), would Gauss Law be still true? (c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the line of force passing through that point? (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

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(e) We know that electric field is discontinuous across the surface of a charged conductor, is electric potential also discontinuous there? (f) What meaning would you give to the capacity of a single conductor? (g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say mica (=6)?

• Solution: (a) No, because charge distribution on the spheres will not be uniform. (b) Gauss Law will not be true. (c) Not necessarily (True only if the line of force is a straight line). The line of force gives the direction of acceleration, not that of velocity, in general. (d) Zero, no matter what the shape of the complete orbit. (e) No, potential is continuous. (f) A single conductor is a capacitor with one of the 'plates' at infinity. (g) Water molecules have permanent dipole moments.

31. (a) Calculate the potential at a point P due to a charge of located 9 cm away.

(b) Hence obtain the work done in bringing a charge of from infinity to the point P. Does the answer depend on the path along which the charge is brought?

• Solution:

a)

(b) W=qV=

No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

32. Two charges are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

• Solution: Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin [refer the figure]

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Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x<0.) If x lies between O and A, we have

where x is in cm. That is,

which gives x = 9 cm.

If x lies on the extended line OA, the required condition is

which gives x = 45 cm

Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

33. Figures (a) and (b) show the field lines of a positive and negative point charge

respectively.

(a) Give the signs of the potential difference VP – vQ; VB-VA.

(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.

(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.

(d) Give the sign of the work done by the external agency in moving a small negative charge from B to A.

(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A?

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• Solution:

(a) As V ∞ Thus, ( ) positive. Also VB is less negative than VA.

Thus, or (VB-VA) is positive. (b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of the potential energy difference of a small negative charge between Q and P is positive. (c) Similarly, (P.E)A > (P.E)B and hence sign of potential energy differences is positive. In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative. (d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive. (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

34.

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Four charges are arranged at the corners of a square ACD of side d, as shown in the figure. Find the work required to put together this arrangement (b) A charge q0 is brought to the center E of the square, the four charges being held fixed at its corners. How much extra work is needed for this?

• Solution: (a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps: (1) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero. (2) Work needed to bring –q to B when +q is at A. This is given by (charge at B) ×

(electrostatic potential at B due to charge +q at A) = (3) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at

C) × (electrostatic potential at C due to charges at A and B) = (4) Work needed to bring –q to D when +q t A, -q at B, and +q at C. This is given by (charge at C) × (electrostatic potential at C due to charges at A, B and C)

=

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Add the work done in steps (i) , (ii), (iii) and (iv). The total work required is

=

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.

(b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q0 × (electrostatic potential at E due to charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence no work is required to bring any charge to point E.

35. (a) Determine the electrostatic potential energy of a system consisting of two

charges 7 µµµµ C and –2 µµµµ C (and no external field) placed at (-9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r2) ; A = 9 ×××× 105 Cm-2. What would the electrostatic energy of the configuration be?

• Solution:

(a) U = (b) W = U2 -U1 = 0- U = 0 - (-0.7) = 0.7J (c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,

Q1 V(r1) +q2 V(r2) = And the net electrostatic energy is

Q1 V(r1) +q2 V(r2) + = 70 - 20 - 0.7 = 49.3 J.

36. A molecule of a substance has a permanent electric dipole moment of magnitude 10-

29 C m. A mole of this substance is polarized by applying a strong electrostatic field of magnitude 106 V m-1. The direction of the field is suddenly changed by an angle of 60 °°°° . Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100 % polarization of the sample.

• Solution: Here, dipole moment of each molecules = 10-29 C m. As 1 mole of the substance contains 6 × 1023 molecules, Total dipole moment of all the molecules, p = 6 × 1023 × 10-29 C m = 6 × 10-6 C m

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Initial potential energy, Ui = -pE cos θ = -6 × 10-6 × 106 cos 0° = -6J Final potential energy = Uf = -6 × 10-6 × 106 cos 60° = -3J

Change in potential energy = -3J – (-6J) = 3J

So, there is loss in potential energy. This must be the energy released by the substance in te form of heat in aligning its dipoles.

37. (a) A comb run through one’s dry hair attracts small bits of paper. Why? What

happens if the hair is wet on a rainy day? (b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary? (c) Vechicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why? (d) A bird perches on a bare high power line, and nothing happens to the ird. A man standing on the ground touches the same line and gets a fatal shock. Why?

• Solution: (a) This is because the comb gets charged by friction. The molecules in the paper gets polarized by the charged comb, resulting in a net force of attraction. If the hair is wet, or if it is rainy day, friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper. (b) To enable them to conduct charge to the ground; as too much of static electricity accumulated may result in spark and result in fire. (c) Reason similar to (b) (d) Current passes only when there is difference in potential.

38. A network of four 10 µµµµ F capacitors is connected to a 500 V supply, as shown in the

figure. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor.

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• Solution: (a) In the given network, C1, C2, C3 are connected in series. The effective capacitance C' of these three capacitors is given by

For C1 = C2 = C3 = 10 µ F, C' = (10/3) µ F. The network has C' and C4 connected in parallel. Thus, the equivalent capacitance C of the network is

C = C' + C4 =

(b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 e Q'. Now since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3, we have

Also, Q'/C4 = 500 V This gives for the given value of the capacitances, Q = 500 V × (10/3) µ F = 1.7 × 10-3 C

Q' = 500 V × 10 µ F = 5.0 × 10-3 C.

39. (a) A 900 pF capacitor is charged by 100 V battery as in the figure. How much

electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

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• Solution: (a) The charge on the capacitor is Q = CV = 900 × 10-12 F × 100 V = 9 × 10-8 C The energy stored in the capacitor = (1/2) CV2 = (1/2) QV = (1/2) × 9 × 10-8 C × 100 V = 4.5 × 10-6 J

(b) In the steady situation, the two capacitors have their positive plates at the same potential. Let the common potential difference e V’. The charge on each capacitor is Q’ = CV’. By charge conservation, Q’ = Q/2. This implies V’ = V/2. The total energy of the system is = 2 × (1/2) Q’V’ =

(1/4) QV = 2.25 × 10-6 J

Thus in going from (a) to (b), through no charge is lost; the final energy is only half the initial energy.

There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.

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chapter 2. electrostatic potential and … 12th/class...3. two charges 2µµµµc and -2µµµc are...

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