PH 108 : Electricity & Magnetism Answer Booklet

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PH 108 : Electricity & Magnetism Answer Booklet Rwitaban Goswami 2019-20 1

Transcript of PH 108 : Electricity & Magnetism Answer Booklet

Page 1: PH 108 : Electricity & Magnetism Answer Booklet

PH 108 : Electricity & MagnetismAnswer Booklet

Rwitaban Goswami

2019-20

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Page 2: PH 108 : Electricity & Magnetism Answer Booklet

PH 108 : Electricity & Magnetism Contents Rwitaban Goswami

Contents

1 Tutorial Sheet 3

2 Tutorial Sheet 6

3 Tutorial Sheet 9

4 Tutorial Sheet 13

5 Tutorial Sheet 19

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PH 108 : Electricity & Magnetism Tutorial Sheet: 1 Rwitaban Goswami

1 Tutorial Sheet

1.1. Calculate the curl and divergence of the following vector functions. If the curl turns out to bezero, construct a scalar function φ of which the vector field is the gradient:

(a) Fx = x+ y ; Fy = −x+ y ; Fz = −2z

Solution:

Curl: ∇× ~F = −2k

Div: ∇ · ~F = 0

(b) Gx = 2y ; Gy = 2x+ 3z ; Gz = 3y

Solution:

Curl: ∇× ~G = ~0

Div: ∇ · ~G = 0

φ(x, y, z) = 2xy + 3yz + C

(c) Hx = x2 − z2 ; Hy = 2 ; Hz = 2xz

Solution:

Curl: ∇× ~H = −2z+ 2yk

Div: ∇ · ~H = 4x

1.2. A ship sails from the southernmost point of India (6.75oN, 93.84oE) to the southernmost pointof Africa (34.5oS, 20.00oE) following the shortest possible path.

(a) Given that the radius of the earth is 6400 km, what is the distance it has covered?

Solution: 9017 km

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(b) If instead of sailing, one had travelled in an aeroplane - by what percentage would theshortest possible distance change?

Solution: = 0.15%

1.3. A force defined by ~F = A(y2i + 2x2j) is exerted on a particle which is initially at the originof the co-ordinate system. A is a positive constant. We transport the particle on a triangularpath defined by the points (0,0,0), (1,0,0), (1,1,0) in the counterclockwise direction.

(a) How much work does the force do when the particle travels around the path? Is this aconservative force?

Solution: A, Non conservative

(b) The particle is placed at rest right at the origin. Is this a stable situation? Give anyargument (mathematical, physical, intuitive) to justify the stability (or instability) of thissituation.

Solution: The particle is not in stable equilibrium.

1.4. The area bounded by the curve r = 2R cos θ has a surface charge density σ(r, θ) = σ0rR

sin4 θ.What is the total amount of charge?

Solution:

32σ0R2

105

1.5. Consider a sphere of unit radius centered at the origin. Show that the area of the surfaceenclosed between θ = 0 and θ = α is given by 2π(1− cosα)

Solution: Proof in Solution Booklet

1.6. For a circle given by the equation, r = 2R cos θ, where R is the radius of the circle and (r, θ)are the polar coordinates. Calculate the following

(a) area of the circle

Solution:

πR2

(b) centroid of the semi-circular area in the first quadrant

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Solution:

R ı+4R

(c) length and centroid of the semi-circular arc in the first quadrant

Solution:

Length = πR

Centroid = R ı+2R

π

(d) area common to the given circle and the circle given by r = R

Solution:

2πR2

3−√

3R2

2

1.7. Consider the section of a cone described by the equation z2 = x2 + y2, between the planes z = 1and z = 2. Determine the volume of this part using

(a) spherical polar co-ordinates,

Solution:

3

(b) cylindrical co-ordinates.

Solution:

3

1.8. Suppose that the height of Sameer Hills (in feet) is given by

h(x, y) = 10(2xy − 3x2 − 4y2 + 14x+ 10y + 40),

where x is the distance (in km) east, y the distance north of Hostel 16

(a) Where is the top of Sameer Hills located, and how high is it?

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Solution: So the top of Sameer hills is located 3 km north and 2 km east of Hostel16, and it is at a height of 710 feet

(b) How steep is the slope (in feet per km) at a point 1 km north and 1 km east of Hostel 16?In what direction is the slope steepest, at that point?

Solution: So the steepness of the slope at (1, 1) is 20√

29 and the slope is steepest at1√29

(5 ı+ 2 )

2 Tutorial Sheet

2.1. Compute the divergence of the vector field given by:

~v = r cos θ r + r sin θ θ + r sin θ cosφ φ

Check the divergence theorem for this using the volume of an inverted hemisphere of radius R,resting on the xy plane and centered at the origin.

Solution:

∇ · ~v = 5 cos θ − sinφ

Flux =5πR3

3

2.2. Check the divergence theorem for the function

~v = r2cos θ r + r2cosφ θ − r2cos θsinφ φ

using the volume of one octant of a sphere of radius R.

Solution:

∇ · ~v = 4r cos θ

Flux =πR4

4

2.3. Compute the unit normal vector n to the ellipsoidal surfaces defined by constant values of

Φ(x, y, z) = V

(x2

a2+y2

b2+z2

c2

). What is n when a = b = c?

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Solution:

n =xa2ı+ y

b2+ z

c2k√

x2

a4+ y2

c4+ z2

c4

When a = b = c, n = r

2.4. A vector field is given by~v = ay i+ bx j

where a, b are constants.

(a) Find the line integral of this field over a circular path of radius R, lying in the xy planeand centered at the origin using (i) Plane polar, (ii) Cartesian

Solution:

(b− a)πR2

(b) Imagine a right circular cylinder of length L with its axis parallel to the z axis standingon the circle. Use cylindrical co-ordinate system to show that the Stokes theorem is validover its surface.

Solution: Validity of Stokes in Solution Booklet

2.5. Express the following derivatives in terms of linear combinations of the unit vectors of thespherical polar co-ordinate, r, θ, φ :

∂r

∂θ,

∂r

∂φ,

∂θ

∂θ,

∂θ

∂φ,∂φ

∂φ

Solution:

∂r

∂θ= θ,

∂θ

∂θ= −r, ∂r

∂φ= sin θ φ,

∂θ

∂φ= cos θ φ,

∂φ

∂φ= − sin θ r − cos θ θ

2.6. The gradient operator ∇ behaves like a vector in “ some sense”. For example, divergence of acurl (∇.∇× ~A = 0) for any ~A, may suggest that it is just like ~A. ~B × ~C being zero if any two

vectors are equal. Prove that ∇×∇× ~F = ∇(∇. ~F )−∇2 ~F . To what extent does this look like

the well known expansion of ~A× ~B × ~C ?

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Solution: Proof in Solution Booklet

2.7. As a more involved example, show that the operator L = −i~r × ∇ where (i =√−1) satisfies

L × Lf = iLf where f is an arbitrary test function. (Notice that the cross product of anoperator with itself does not necessarily vanish, can you see why?)

Solution: Proof in Solution Booklet

2.8. A vector field is given by~v = xyi + 2yzj + 3zk

Verify the validity of Stoke’s theorem, using a triangular area with vertices (0, 0, 0), (0, 2, 0) and(0, 0, 2).

Solution:

Line integral = −8

3

Validity of Stokes in Solution Booklet

2.9. Although the gradient, divergence and curl theorems are the fundamental integral theorems ofvector calculus, it is possible to derive a number of corollaries from them. Show that:

(a)´V

(∇T ) dV =¸ST d~S [Hint: Let ~v = ~c T where ~c is a constant, in the divergence theorem]

Solution: Proof in Solution Booklet

(b)´V

(∇ × ~v) dV = −¸S~v × d~S [Hint: Replace ~v by ~v × ~c, where ~c is a constant in the

divergence theorem ]

Solution: Proof in Solution Booklet

2.10. Consider an arbitrary surface S and it’s boundary C

(a) Calculate 1V

¸S~r · d~S, if S is closed and it encloses volume V

Solution: 3

(b) Calculate¸Sd~S, if S is closed and it encloses volume V [Hint : Use 2.9a]

Solution: 0

(c) Show that´Sd~S = 1

2

¸C~r × d~C

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Solution: Proof in Solution Booklet

3 Tutorial Sheet

3.1. A small ball with a positive charge +q hangs by an insulating thread.Holding this ball vertical,a second ball having charge +q is kept at a distance a along the horizontal direction. There arean infinite number of points where a third ball with charge +2q may be positioned so that thefirst ball continues to remain vertical when released. Find the equation of the curve describingthese points.

Solution:

y = +

√(2a2x)

23 − x2

3.2. A semi-infinte slab of thickness t has a uniform charge density ρ distributed in its volume. Findthe electric field intensity at a distance z from the median plane of the slab. Consider values ofz both inside and outside the slab.

z

~E

Solution:

Ez =

{ρt4ε0

if |z| > t2

ρz4ε0

if |z| ≤ t2

Ehorizontal →∞

3.3. A thin annular disc of inner radius a and outer radius b carries a uniform charge density σ.Determine the electric field intensity at a point on the z-axis (the axis of symmetry). Using thisresult determine the field due to an infinite sheet containing a charge density σ.

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Solution:

~Eannular disc =σz0

2ε0

(1√

a2 + z20

− 1√b2 + z2

0

)z

~Einfinite sheet =σ

2ε0z

3.4. Which one of the following is possible expression for an electrostatic field? For the right expres-sion, find a potential which determines this field with the origin as the reference. (here A is aconstant having appropriate dimensions)

(a) ~E = A(xyz2ı+ 2xz− 3yzk)

Solution: This is not a valid electrostatic field

(b) ~E = A(

(3xz2 + y2)ı+ 2xy+ 3x2zk)

Solution: This is a possible electrostatic field.

phi = A

(−3

2x2z2 − xy2

)+ C

3.5. The charge distribution of the hydrogen atom produces an electric field

~E = C[1− e−αr] rr2

where C and α are constants. Find the net charge within a sphere of radius r = 1α

.

Solution:4πε0C

(1− e−1

)

3.6. Show that the maximum value of the electric field |E| for points on the axis of a uniform ringof radius R with total charge q occurs at x = ± R√

2. If an electron is placed at the centre of

the ring and then displaced by a small amount x(x � R) along the axis, show that it wouldexecute simple harmonic oscillations. Determine the frequency of oscillations.

Solution: The oscillation frequency is√

qe4πε0R3me

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3.7. A charged semicircular ring of radius R extending from θ = 0 to θ = π lies in the xy plane,centered at origin. If the charge distribution on the ring is λ0 sin θ, compute the electric fieldintensity at P (0, 0, z).

Solution:

λ0

4πε0

R√R2 + z2

0

3

(−π

2R y + 2z0 z

)

3.8. A hemisphere of radius R has z = 0 as its equatorial plane and lies entirely in the region z ≥ 0.The hemisphere has a uniform volume charge density ρ. Determine the field at the center ofthe hemisphere.

Solution:

−ρR4ε0

k

3.9. A sphere has a uniform volume charge density everywhere except inside an off-centre sphericalcavity within. Show that the field inside the cavity is uniform.

Solution: Proof in Solution Booklet

3.10. Two infinite sheets of planes intersect at right angles. The sheets carry charge densities +σand −σ. Find the magnitude and direction of electric field everywhere and sketch the electricfield lines.

Solution:

∣∣∣ ~E∣∣∣ =σ√2ε0

Direction of field lines is ‖ to xz plane and at 450 to xy plane

3.11. Consider a point charge q moving with speed v along the z-axis (i.e. θ = 0 direction). Theelectric field due to such a moving charge is given by (do not worry how it is derived, you donot need to know that to solve this)

~E(r, θ) =q

4πε0

(1− β2

) 1(1− β2sin2θ

)3/2

r

r2

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β =v

c, where c is the speed of light in vacuum. Notice that the field remains radial at all points.

You can use the substitution tanα =β√

1− β2cos θ to evaluate the integral.

(a) Compared to the usual Coulomb field, at which location does it become stronger/weaker?

Solution: Er is stronger than the Er for a stationary charge for

sin θ >

√1− (1− β2)

23

β2

or

sin θ < −

√1− (1− β2)

23

β2

(b) Calculate the flux of the electric field through the curved surface of the top hemispherebetween (θ = 0 to θ = π/2) and the bottom hemisphere (θ = π/2 to θ = π). Which islarger? What is the sum total of the flux through the two hemispheres?

Solution:

Flux through top =q

2ε0

Flux through bottom =q

2ε0

Total flux is qε0

(c) Calculate the flux through the forward cone between (θ = 0 to θ = π/4) and compare itwith the value for v = 0. Does this increase or decrease?

Solution:

Flux =q

2ε0

(1− 1√

2− β2

)

Flux decreases as β increases

For interested students, this field is actually consistent with special relativity. Youcan notice that if you draw the field lines of a stationary charge, and lorentz contractthat image, then you will get the field lines of the moving charge

3.12. Consider a charge distribution that produces a potential everywhere in space described by,

V (r, θ, φ) =1

4πε0

A

r

(1 +

B

rcos θ

)where A and B are two constants.

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(a) Calculate the electric field ~E(r, θ, φ) everywhere.

Solution:

~E = − A

4πε0

((1

r2+

2B cos θ

r3

)r +

B sin θ

r3θ

)

(b) Calculate the total charge Q of the charge distribution that produces the given potential.

Solution:

Q = A

You can see that the potential is actually of a point charge and a dipole, so the total chargemust be due to the point charge only, so our answer does make sense

3.13. Consider a sphere of radius R, containing a uniform volume charge density ρ0, centered at theorigin.

(a) Assuming the potential at infinity to be zero, calculate the potential at the point (R, 0, 0)and at the origin.

Solution: At origin, V = ρR2

2ε0

At (R, 0, 0), V = ρR2

3ε0

(b) A thin slice of charge, of thickness t with t << R, is cut out from the equitorial xy planeof the above sphere. Calculate the change in the potential at the origin and at the point(R, 0, 0).

Solution:

Vorigin = −ρtR2ε0

V(R,0,0) = −ρtRπε0

4 Tutorial Sheet

4.1. (a) A sphere of radius R carries a charge density ρ(~r) = kr (where k is some constant). Findthe electrostatic energy of this charge configuration.

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Solution:

πk2R7

7ε0

(b) You can think about the electrostatic energy in two ways. Either

Total Electrostatic energy U =ε02

˚all space

~E · ~E dV

or we can say that

Electrostatic energy density =ε02~E · ~E

Are these two statements physically equivalent? Does the second way imply some additionalphysical property compared to the first?

Solution: These two statements are not physically equivalent.In the first case we are looking at energy on a global scale. So in the context ofconservation of energy, it only demands global conservation of energy. It allows for Uamount of energy to dissappear from here and U amount of energy to simultaneouslyappear there. But this would violate local conservation of energy which the secondcase implies.So the second case gives a definite location to the the energy, and imposes the additionalconstraint of local conservation.

In the words of Richard Feynman, “You realize that our early statement of the principleof the conservation of energy is still perfectly all right if some energy disappears atone place and appears somewhere else far away without anything passing (that is,without any special phenomena occurring) in the space between. We are, therefore,now discussing an extension of the idea of the conservation of energy. We might callit a principle of the local conservation of energy. Such a principle would say that theenergy in any given volume changes only by the amount that flows into or out of thevolume. It is indeed possible that energy is conserved locally in such a way. If it is, wewould have a much more detailed law than the simple statement of the conservation oftotal energy. It does turn out that in nature energy is conserved locally. We can findformulas for where the energy is located and how it travels from place to place.”

(c) There exists two ways of writing the total electrostatic energy in a system

U =ε02

˚V

~E · ~E dV (1)

U =1

2

˚Vρφ dV (2)

where φ is the electrostatic potential and ρ is the volume charge density. Now what dothese two ways imply about the location of Electrostatic energy? Do they imply differentlocations? If so can you say anything about which one is more “physically true”?

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Solution: The first case implies that electrostatic energy is located in the fieldThe second case implies that electrostatic energy is located in the chargesAt our current knowledge of physics (i.e. classical mechanics, quantum mechanics,electrodynamics), it is not possible to say anything about which might be “truer”.No physical phenomenon or their prediction will change if we assume one or the other.But, if we introduce General Relativity to the mix, it can indeed be ascertained thatwhere the energy is located, because energy contributes to the bending of spacetime,and the location of this energy affects how spacetime is bent.It has thus been found that the electrostatic energy is located in the field

In the words of Richard Feynman, “It is interesting that there seems to be no uniqueway to resolve the indefiniteness in the location of the field energy. It is sometimesclaimed that this problem can be resolved by using the theory of gravitation in thefollowing argument. In the theory of gravity, all energy is the source of gravitationalattraction. Therefore the energy density of electricity must be located properly if weare to know in which direction the gravity force acts. As yet, however, no one has donesuch a delicate experiment that the precise location of the gravitational influence ofelectromagnetic fields could be determined. That electromagnetic fields alone can bethe source of gravitational force is an idea it is hard to do without. It has, in fact,been observed that light is deflected as it passes near the sun-we could say that the sunpulls the light down toward it. Do you not want to allow that the light pulls equallyon the sun? Anyway, everyone always accepts the simple expressions we have foundfor the location of electromagnetic energy and its flow. And although sometimes theresults obtained from using them seem strange, nobody has ever found anything wrongwith themthat is, no disagreement with experiment. So we will follow the rest of theworld-besides, we believe that it is probably perfectly right. ”

4.2. The potential takes the constant value φ0 on the closed surface S which bounds a volume V .The total charge inside V is Q. There is no charge anywhere else. Show that the electrostaticenergy contained in the space outside of S is UE(out) = Qφ0

2

Solution: Proof in Solution Booklet

4.3. Let the space between the two concentric spheres of radii a and R (R ≥ a) be filled uniformlywith charge

(a) Calculate the total energy UE in terms of the total charge Q and the variable x = aR

. Checkthe a = 0 and a = R limits.

Solution:

UE =Q2

8πε0R

(1 +

15− x3 + 9

5x5 − x6

(1− x3)2

)

(b) Minimize UE with respect to x (keeping the total charge Q). Identify the physical systemwhich achieves the minimum you find.

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Solution: x = 1Uniformly charged shell

4.4. Consider an idealised vacuum diode consisting of two large parallel plates (like a capacitor).The distance between them is D, the plates are kept at V (x = 0) = 0 and V (x = D) = V0. HereV0 is positive. The plate at V = 0 is then heated, so that it starts emitting electrons which areattracted towards the other plate. You need to calculate the steady state current (I) - voltage(V0) relation (called the Child-Langmuir law).

(a) The space between the two plates will be filled with electrons so the charge density thereis not zero. The velocity of the charged particles depend on the position x, however inthe steady state the charge density remains time independent and the current is constant.Under these conditions, formulate the Poisson’s equations such that V (x) and x are theonly variables.

Solution: d2Vdx2

= − 1ε0ρ

(b) Give a physical argument to show that the steady state would be reached when the electricfield at the surface of the heated plate at x = 0 is zero.

Solution: As the temperature is increased, more and more electrons are emmittedfrom the cathode, so much so that after a point there is so much buildup of the cloudof electrons that it starts to repel the incoming electrons from cathode.This is the steady state, where the space charge makes the electric field at cathodeexactly 0.The Child-Langmuir law is manifest whenever there is space charge limited current

(c) Use this as your boundary condition to solve the differential equation fully. Establish therelation I ∝ V0

3/2

Solution:

I = KV32

0 , where K =4ε0A

9d2

√2q

m

4.5. A thin wire carrying line charge density λ extends along the z-axis from z = −d to z = d

(a) By integration over the charge distribution, find out the potential at any general point(x, 0, z) in the xz plane

Solution:

λ

4πε0ln

((z + d) +

√x2 + (z + d)2

(z − d) +√x2 + (z − d)2

)

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(b) Find out the location of a point on the x-axis where the potential is the same as that thepoint (0, 0, 2d)

Solution:

x = ±√

3d

(c) Check the distances of these equipotential points from the end points of the charge distri-bution and find the nature (shape) of the equipotential curve on which the two points andreside.

Solution: These points form an ellipse with the end points of the charge distributionas the focii

4.6. Consider an infinite chain of point charges ±q (with alternating signs), strung out along the xaxis, each a distance a from its nearest neighbors. Find the work done per particle requiredto assemble this system. (If this comes out to be −αq

2

4πε0afor some dimensionless number α; your

problem is to determine α. It is known as the Madelung constant.)

Solution: α = ln(2)

4.7. Find the potential on the axis of a uniformly charged solid cylinder, a distance z from thecenter. The length of the cylinder L, its radius R, and the charge density is ρ. Use your resultto calculate the electric field at this point. (Assume that z > L

2)

Solution:

V =ρ

4ε0

(z +L

2

)√R2 +

(z +

L

2

)2

−(z − L

2

)√R2 +

(z − L

2

)2

4ε0

R2 ln

z + L2

+√R2 +

(z + L

2

)2

z − L2

+√R2 +

(z − L

2

)2

− 2zL

~E = − ρ

4ε0

2

√R2 +

(z +

L

2

)2

− 2

√R2 +

(z − L

2

)2

− 2L

z

4.8. Two spherical cavities of radii a and b, are hollowed out from the interior of a (neutral) con-ducting sphere of radius R. At the center of each cavity a point charge is placed: qa andqb.

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R

qaa

qbb

(a) Find the surface charge densities σa, σb, σR

Solution:

σa = − qa4πa2

σb = − qb4πb2

σR =qa + qb4πR2

(b) What is the field outside the conductor?

Solution:~Eout =

1

4πε0

qa + qbr2

r

(c) What is the field within each cavity?

Solution:

~Ea =1

4πε0

qar2a

ra

~Eb =1

4πε0

qbr2b

rb

(d) What is the force on qa and qb?

Solution: Zero

(e) Which of these answers would change if a third charge, qc, were brought near the conductor?

Solution: σR changes but not σa or σb~Eoutside changes but not ~Ea or ~EbForce on qa and qb are still zero

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5 Tutorial Sheet

5.1. A point charge +q is placed at a distance d from the centre of a conducting sphere of radius R(d > R). Show that if the sphere is grounded, the ratio of the charge on the part of the sphere

visible from +q to that on the rest is√

d+Rd−R .

Solution: Proof in Solution Booklet

5.2. Consider a conducting sphere A which is initially uncharged. Another conducting sphere B isgiven a charge +Q, brought into contact with A and then moved far away. The charge on Bis then increased to its original value +Q and again brought into contact with A. Show that if

this process is repeated many times, the charge on A will tend to the limitQq

Q− q, where q is

the charge acquired by A after its first contact with B.

Solution: Proof in Solution Booklet

5.3. Two infinite conducting plates (both grounded and perpendicular to the x − y plane) meet atan angle of 60◦. A point charge +q in the xy plane has plane polar coordinates (a, 20◦). Findall the image charges and their positions in polar coordinates.

Solution:

(a,−20◦)

(a, 100◦)

(a,−100◦)

(a, 160◦)

(a,−160◦)

5.4. Two infinite parallel grounded conducting plates are held a distance a apart. A point charge qis placed between them, at a distance x from one plate. Find the force on q. Check that youranswer is correct for the special cases a→∞ and x = a

2.

Solution:

F = − q2

16πε0

{1

x2− 4ax

∞∑n=1

n

(n2a2 − x2)2

}

When a→∞F =

q2

4πε0

1

(2x)2

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PH 108 : Electricity & Magnetism Tutorial Sheet: 5 Rwitaban Goswami

which is the same force as the case for a single plane.If x = a

2

F = 0

5.5. A point charge q of mass m is released from rest at a distance d from an infinite groundedconducting plane. How long will it take for the charge to hit the plane?

Solution:

T =πd

q(2πε0md)

12

5.6. A grounded conductor has a cavity of radius R inside it. A point charge +q is placed at adistance r (r < R) from the centre of the cavity.

(a) Find the electrostatic force experienced by the charge +q and show that for r << R, thisforce is proportional to r.

Solution:

F =1

4πε0R3

q2r(1− r2

R2

)2

For r � R

F =q2

4πε0R3r

(b) Find the expression for the charge density at two diametrically opposite points which arecollinear with q, on the surface of the cavity.

Solution: Let us first look at the point closer to +q

σ = − q

4πR

R + r

(R− r)2

Let us now look at the point furthur from +q

σ = − q

4πR

R− r(R + r)2

5.7. A circular ring of radius b has a uniform linear charge density λ and is placed on the xy planewith the centre at the origin.

(a) Calculate potential at point (0, 0, z)

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Page 21: PH 108 : Electricity & Magnetism Answer Booklet

PH 108 : Electricity & Magnetism Tutorial Sheet: 5 Rwitaban Goswami

Solution:

φ =λ

2ε0

b√b2 + z2

(b) A conducting sphere of radius a with a = b/2 is placed at the centre of the ring, with theircentres coincident. Calculate the percentage change in the potential at the point (0, 0, 2a)when the conducting sphere is grounded.

Solution:

%− change =14

12√

2− 1√

17+ 1

4

= 69%

5.8. A conducting sphere (or a shell) of radius R has a charge Q.

(a) Show that the force of repulsion between the two hemispheres is Q2/32πε0R2.

Solution: Proof in Solution Booklet

(b) Now suppose one has a solid sphere of radius R with charge Q distributed uniformly overits volume. What will be the force of repulsion between the two hemispheres?

Solution:

3Q2

64πε0R2

(c) In which case(a or b) is the force of repulsion larger?

Solution: As we can see, 364> 1

32, so the force of repulsion is lesser for the conductor.

This is what we should expect, since in a conductor the charges always distribute sothat the repulsion is minimized

5.9. A spherical soap bubble carries a charge Q. Its surface tension is T . The definition of surfacetension is that it is extra energy per unit area of the surface. So one needs to do some workto increase the surface area of a bubble. Identify all the sources of energy and then answer thefollowing:

(a) Show that excess pressure inside an uncharged soap bubble of radius R should be ∆P = 4TR

.If a soap film has T = 0.03Nm−1 and a radius of 1cm, how much would be the excesspressure? (FYI : the surface tension of water at 20◦C is T = 0.072Nm−1)

Solution: Proof in Solution Booklet

(b) Now, if the bubble has Q charge in it - should ∆P increase or decrease? Give a qualitativeargument first.

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Page 22: PH 108 : Electricity & Magnetism Answer Booklet

PH 108 : Electricity & Magnetism Tutorial Sheet: 5 Rwitaban Goswami

Solution: If the charge Q is put on the bubble, each hemisphere will attract the othera little lesser, because there is an extra electrostatic repulsive force. So the excesspressure ∆P required inside the bubble to balance that force will be lesser. So ∆Pdecreases

(c) Now work out a quantitative expression for ∆P . How much charge is required to make thebubble burst?

Solution: ∆P = 4TR− Q2

32π2ε0R4

The bubble will burst when Q =√

64π2Tε0R3

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