Applied Physics PDF - 24PH101T - August 3, 2024

Applied Physics Course Code: 24PH101T Applied Physics August 3, 2024 1 / 66 Outline 1 Coulomb’s Law and Electric Field 5 Electric Potential 2 Continuous Charge Distribution 6 Work and Energy in 3 Gauss’s Law Electrostatics 4 Curl of...

Applied Physics Course Code: 24PH101T Applied Physics August 3, 2024 1 / 66 Outline 1 Coulomb’s Law and Electric Field 5 Electric Potential 2 Continuous Charge Distribution 6 Work and Energy in 3 Gauss’s Law Electrostatics 4 Curl of Electric Field 7 Problems (Electrostatics) Applied Physics August 3, 2024 2 / 66 Coulomb’s Law – (1) Coulomb’s Law The force on charge Q due to a single point charge q (both are at rest) separated by a distance r. 1 qQ F⃗ = r 4πϵ0 r 2 c ϵ0 = 8.85 × 10−12 C 2 N −1 m2. ϵ0 is called permittivity of free space. The force points along the line from q to Q, it it is repulsive. The force points along the line from Q to q, it it is attractive. Applied Physics August 3, 2024 3 / 66 Coulomb’s Law – (2) Fundamental Problem There are charges q1 , q2 , q3 , etc (called source charrges). What is the force exerted by these charges on a test charge Q? First assumption All charges are stationary. Principle of Superposition The interaction between any two charges is completely unaffected by the presence of others. Applied Physics August 3, 2024 4 / 66 Coulomb’s Law – (3) This means that to determine the force on Q, one can first compute the force F⃗1 , due to q1 alone. Then compute the force F⃗2 , due to q2 alone; and so on. Finally, the vector sum of all these individual forces: F⃗ = F⃗1 + F⃗2 + F⃗3 +... to calculate the total force. Coulomb’s law and the principle of superposition constitute the physical input for electrostatics – the rest, except for some special properties of matter, is mathematical elaboration of these fundamental rules. Applied Physics August 3, 2024 5 / 66 Example 1 – (1) Question Twelve equal charges, q, are situated at the corners of a regular 12-sided polygon (for instance, one on each numeral of a clock face). Calculate the net force on a test charge Q at the center. For charge present at +ve Z For charge present at −ve Z 1 qQ 1 qQ F⃗1 = − zb F⃗2 = zb 4πϵ0 r 2 4πϵ0 r 2 Equating these two, the net force is zero. Similarly, for all the pair of charges. Hence, the net force on test charge Q at the center is zero. Applied Physics August 3, 2024 6 / 66 Example 1 – (2) Question Suppose one of the 12 q’s is removed. What is the force on Q? Explain your reasoning carefully. Assume the one at −ve Z is removed (“6 o’clock”). The net force on the charge Q due to charges at +ve Z and −ve Z is now not balanced. 1 qQ F⃗net = − zb 4πϵ0 r 2 For all the other pairs, the forces between opposite charges, all cancel out. Applied Physics August 3, 2024 7 / 66 The Electric Field – (1) If we have several point charges q1 , q2 ,... , qn at distances r 1 , r 2 ,... , r n from Q, the total force on Q is, F⃗ = F⃗1 + F⃗2 +... + F⃗n   Q q1 q2 qn F⃗ = r 1+ r 2 +... + rn r 2 r r d 2 d 2 d 4πϵ0 1 2 n Applied Physics August 3, 2024 8 / 66 The Electric Field – (2) Force can be, F⃗ = Q E⃗ It makes no reference to the test charge Q. Where E⃗ is, The electric field is a vector n quantity that varies from 1 X qi E⃗ = rc point to point and is 4πϵ0 r 2i i i=1 determined by the configuration of source charges. E⃗ is called the electric field of the source charges qi ’s. Physically, E⃗ (⃗r ) is the force per unit charge that would be It is a function of position (⃗r ), exerted on a test charge, if because the separation vectors you were to place one at P. r⃗ i depend on the location of the field point P. Applied Physics August 3, 2024 9 / 66 The Electric Field – (3) What is an electric field? It is encouraged to think of the field as a “real” physical entity, filling the space around electric charges. Maxwell himself came to believe that electric and magnetic fields are stresses and strains in an invisible primordial jellylike “ether”. Special relativity has forced us to abandon the notion of ether, and with it Maxwell’s mechanical interpretation of electromagnetic fields. It is even possible, though cumbersome, to formulate classical electrodynamics as an “action-at-a-distance” theory, and dispense with the field concept altogether. I can’t tell you, then, what a field is—only how to calculate it and what it can do for you once you’ve got it. Applied Physics August 3, 2024 10 / 66 Example 2 – (1) Question Find the electric field a distance z above the midpoint between two equal charges (q), a distance d apart In such problems, remember, r 1 = r 2 = r but dr 1 ̸= c r Electric Field due to charge at − d2 Electric Field due to charge at + d2 1 q 1 q E⃗1 = r1 E⃗2 = r2 4πϵ0 r 2 4πϵ0 r 2 d d Applied Physics August 3, 2024 11 / 66 Example 2 – (2) The net electric field is, 1 q E⃗ = E⃗1 + E⃗2 = r 1+d r2
4πϵ0 r 2 d It can be easily understood from Figure, r 1 = cosθ zb + sinθ xb d r 2 = cosθ zb − sinθ xb d 1 2q cosθ ∴ E⃗ = zb 4πϵ0 r 2 Applied Physics August 3, 2024 12 / 66 Example 2 – (3) The answer needs to be provided in one of the coordinate system. In this case, Cartesian Coordinates. z The cosθ is defined as, cosθ = and r is r r d2 given by r = z 2 +. 4 The net electric field is, 1 2qz E⃗ =  2 zb 4πϵ0  d2 3 z2 + 4 Applied Physics August 3, 2024 13 / 66 Continuous Charge Distributions – (1) Instead of discrete point charges q1 , q2 , etc. the charge is distributed continuously over some region, in the equation of electric field the sum becomes an integral. Z 1 1 E⃗ (⃗r ) = r dq r 2 c 4πϵ0 Applied Physics August 3, 2024 14 / 66 Continuous Charge Distributions – (2) Different charge distributions 1 If the charge is spread out along a line (Figure b) with charge-per-unit-length λ, then dq = λdl ′ , where dl ′ is an element of length along the line. 2 If the charge is smeared over a surface (Figure c) with charge-per-unit-area σ, then dq = σda′ , where da′ is an element of area on the surface. 3 If the charge fills a volume (Figure d) with charge-per-unit-volume ρ, then dq = ρdτ′ , where dτ′ is an element of volume. dq → λdl ′ ∼ σda′ ∼ ρdτ′ Applied Physics August 3, 2024 15 / 66 Continuous Charge Distributions – (3) Electric Field of line charge, λ(r⃗′ ) Z 1 E⃗ (⃗r ) = r dl ′ r 2 c 4πϵ0 Electric Field of surface charge, σ(r⃗′ ) Z 1 E⃗ (⃗r ) = r da′ r 2 c 4πϵ0 Electric Field of volume charge, ρ(r⃗′ ) Z 1 E⃗ (⃗r ) = r dτ′ r 2 c 4πϵ0 Applied Physics August 3, 2024 16 / 66 Example 3 – (1) Question Find the electric field a distance z above the midpoint of a straight line segment of length 2L that carries a uniform line charge λ. For a infinitesimal line element dl, the required parameters are, xb, r⃗ = −x xb + z zb, ⃗r = z zb, r⃗′ = x √ dl = dx, r = z 2 + x 2. ′   Z λ r⃗′ 1 Z +L λ z zb − x xb 1 E⃗ = E⃗ = r d ⃗l ′ √ dx r2 z2 4πϵ0 +x 2 z2 + x2 c 4πϵ0 −L Applied Physics August 3, 2024 17 / 66 Example 3 – (2) " Z +L Z +L # λ dx x dx E⃗ = zb z 3 − x b 3 4πϵ0 −L (z 2 + x 2 ) 2 −L (z 2 + x 2 ) 2 "  +L #    +L λ x 1 = zb z √ − xb − √ 4πϵ0 z 2 z 2 + x 2 −L z2 + x2 −L 1 2λL = √ zb 4πϵ0 z z 2 + L2 For points very far away from the For very long line segment, line, z >> L, L → ∞, 1 2λL 1 2λ E⃗ ≡ zb E⃗ = zb 4πϵ0 z 2 4πϵ0 z Applied Physics August 3, 2024 18 / 66 Electric Field Lines and Flux – (1) E⃗ of a point charge can be sketched by few representative vectors. (Figure a) Because of 1/r 2 decrease, the length of the vectors keep decreasing as it moves farther away from origin. Another, away is to connect the arrows and form field lines. (Figure b). Prima facie, it looks like the information about the strength of the field is missing. The magnitude of the field is indicated by the density of the field lines. It’s strong near the center where the field lines are close together, and weak farther out. Applied Physics August 3, 2024 19 / 66 Electric Field Lines and Flux – (2) Field lines Such field diagrams are convenient for representing more complicated fields. Ofcourse the number of lines depends on the sharpness of your pencil tip, and how lazy you are... If charge q has 8 field lines, charge 2q should have 16 field lines. Field lines begin on +ve charges and end on -ve charges. Field lines cannot terminate mid air, although they may extend out to infinity. They cannot intersect each other. With all this in mind, it is very easy to sketch the field of point charges of any simple configuration. Applied Physics August 3, 2024 20 / 66 Electric Field Lines and Flux – (3) Flux The flux of E⃗ passing through a surface S, Z ΦE ≡ E⃗ · d⃗a S Flux is a measure of the “number of field lines” passing through S. One can only draw a representative sample of the field lines – the total number would be infinite. But for a given sampling rate the flux is proportional to the number of lines drawn, because the field strength, remember, is proportional to the density of field lines. Applied Physics August 3, 2024 21 / 66 Electric Field Lines and Flux – (4) How it works? The field strength is proportional to the density of field lines (the number of lines per unit area). E⃗ · d⃗a s proportional to the number of lines passing through the infinitesimal area d⃗a. The dot product picks out the component of d⃗a along the direction of E⃗ , as indicated in Figure. It is the area in the plane perpendicular to E⃗ , when calculating the density of field lines. Applied Physics August 3, 2024 22 / 66 Gauss’s Law – (1) The flux through any closed surface is a measure of the total charge inside. Figure (a) The field lines that originate on a positive charge must either pass out through the surface or else terminate on a negative charge inside. Figure (b) A charge outside the surface will contribute nothing to the total flux, since its field lines pass in one side and out the other. Applied Physics August 3, 2024 23 / 66 Gauss’s Law – (2) For a point charge Q at the origin, The total flux passing through 1 Q spherical surface is, E⃗ = rb 4πϵ0 r 2 I Z 1 E⃗ · d⃗a = Q sinθ dθ dϕ Area perpendicular to the surface, 4πϵ0 d⃗a = r 2 sinθ dθ dϕ rb Z sinθ dθ dϕ = 4π Flux passing through a surface d⃗a, 1 Q Total flux is, E⃗ · d⃗a = r · d⃗a) (b 4πϵ0 r 2 I Q 1 E⃗ · d⃗a = = Q sinθ dθ dϕ ϵ0 4πϵ0 Applied Physics August 3, 2024 24 / 66 Gauss’s Law – (3) Suppose there are a bunch of The total flux is, charges scattered about. I n According to the principle of X qi E⃗ · d⃗a = superposition, the total field is the ϵ0 i=1 (vector) sum of all the individual fields: X n For any closed surface, E⃗ = E⃗i I Qenc i=1 E⃗ · d⃗a = ϵ0 The flux through a surface that where, Qenc is total charge encloses them all is, enclosed by the surface. I n I X E⃗ · d⃗a = E⃗i · d⃗a This is the quantitative statement i=1 of Gauss’s law. Applied Physics August 3, 2024 25 / 66 Gauss’s Law – (4) Rewriting Qenc in terms of volume Notice that it all hinges on the charge density ρ, 1/r 2 character of Coulomb’s law; Z without that the crucial Qenc = ρ dτ cancellation of the r ’s would not V take place. Gauss’s law becomes, Gauss’s law is an integral equation, Z   Z but we can easily turn it into a ⃗ ⃗ ∇ · E dτ = ρ dτ differential one, by applying the V V divergence theorem. I Z   Gauss’s law in differential form, ⃗ E · d⃗a = ⃗ · E⃗ dτ ∇ S V ⃗ · E⃗ = ρ ∇ ϵ0 Applied Physics August 3, 2024 26 / 66 Example 4 – (1) Find the field outside a uniformly charged solid sphere of radius R and total charge q. Imagine a spherical surface at radius r > R. This is called a Gaussian surface in the trade. Gauss’s law says that, I Qe nc E⃗ · d⃗a = S ϵ0 In this case Qenc = q. But the quantity we want (E⃗ ) is buried inside the surface integral. Applied Physics August 3, 2024 27 / 66 Example 4 – (2) Applying Gauss’s law, Spherical Symmetry I q For the Gaussian surface, the area E r 2 sinθ dθ dϕ = element, d⃗a = r 2 sinθ dθ dϕ rb S ϵ0 Z π Z 2π 2 q E r sinθ dθ dϕ = The spherical symmetry also shows 0 0 ϵ0 that the electric field is of the q form, 4πr 2 E = ϵ0 E⃗ = E rb Magnitude of E⃗ 1 q ∴ E⃗ · d⃗a = E r 2 sinθ dθ dϕ E= 4πϵ0 r 2 Applied Physics August 3, 2024 28 / 66 Example 5 – (1) A long cylinder carries a charge density that is proportional to the distance from the axis: ρ = ks, for some constant k. Find the electric field inside this cylinder. Consider a Gaussian cylinder of length l and radius s. Charge enclosed within this surface Z Z Z s Z 2π Z 2 Qenc = ρ dτ = ks ′ s ′ ds ′ dϕdz = k s ′2 ds ′ dϕ dz = πls 3 k 0 0 3 Applied Physics August 3, 2024 29 / 66 Example 5 – (2) Cylindrical Symmetry Applying Gauss’s law 2πls 3 k I E⃗ points radially outwards. Es dz dϕ = S 3ϵ0 2π 2πls 3 k Z Z ∴ E⃗ = E sb ∴ Es dz dϕ = 0 3ϵ0 The area element for Gaussian 2πls 3 k ∴ E (2πsl) = surface becomes, 3ϵ0 d⃗a = s dz dϕb s Magnitude of E⃗ ks 2 ∴ E⃗ · d⃗a = Es dz dϕ E= 3ϵ0 Applied Physics August 3, 2024 30 / 66 Example 6 – (1) An infinite plane carries a uniform surface charge σ. Find its electric field. Consider a “Gaussian pillbox,” extending equal distances above and below the plane. If A is the surface area of the lid of pillbox, the charge enclosed in this case is simply Qenc = σA. Symmetry Using this, it shows that E⃗ is pointing away from the plane (upwards for point above, downwards for point below), E⃗ = E nb. Applied Physics August 3, 2024 31 / 66 Example 6 – (2) Area element, Comments: d⃗a = da nb It seems surprising, at first, that the field of an infinite plane is Surface integral yields, independent of how far away the I point is. What about the 1/r 2 in E⃗ · d⃗a = 2E · A Coulomb’s law? S As the point moves farther and Applying Gauss’s law, farther away from the plane, more σA and more charge comes into its 2E · A = ϵ0 “field of view”, and this σ compensates for the diminishing ∴E = influence of any particular piece. 2ϵ0 Applied Physics August 3, 2024 32 / 66 Final answers of Examples Example 4 Example 5 The electric field along with the The electric field along with the direction is, direction is, 1 q ks 2 E⃗ = rb E⃗ = sb 4πϵ0 r 2 3ϵ0 As homework, Example 6 Using the concept from Example The electric field along with the 5, calculate the electric field inside direction is, the uniformly charge solid sphere σ of radius R and total charge q. E⃗ = nb 2ϵ0 Comment: For Example 6, nb is decided by the surface chosen. One side n and the other (downward) it will be −b (upward) it will +b n. Applied Physics August 3, 2024 33 / 66 Curl of Electric Field – (1) For a point charge at origin, the electric field is given by, 1 q E⃗ = rb 4πϵ0 r 2 Using field lines, one can easily conclude that curl of E⃗ is zero. Calculate the line integral of this field from some point a to some other point b. Z b E⃗ · d ⃗l a Applied Physics August 3, 2024 34 / 66 Curl of Electric Field – (2) Solving the integral, In spherical coordinates, Z b   q 1 1 d ⃗l = dr rb + r dθ θb + r sinθ dϕ ϕb E⃗ · d ⃗l = − a 4πϵ0 ra rb 1 q where, ra and rb are distances of E⃗ · d ⃗l = dr points a and b from origin. 4πϵ0 r 2 Integrating, If the integral is for a closed loop, ra = rb , Z b Z b q dr E⃗ · d ⃗l = I a 4πϵ0 a r2 E⃗ · d ⃗l = 0 Applied Physics August 3, 2024 35 / 66 Curl of Electric Field – (3) Applying Stoke’s theorem, I Z   E⃗ · d ⃗l = ⃗ ⃗ ∇ × E · d⃗a S Removing the integral since d⃗a ̸== 0 ⃗ × E⃗ = 0 ∇ Moreover, if there are many charges, the principle of superposition states that the total field is a vector sum of their individual fields, E⃗ = E⃗1 + E⃗2 + E⃗3 +..... Since curl of individual E⃗i ’s is zero, the same is true for total field. Applied Physics August 3, 2024 36 / 66 Example 7 – (1) One of these is an impossible electrostatic field. Which one? 1 E⃗1 = k [xy xb + 2yz yb + 3zx zb] ⃗2 = k y 2 xb + 2xy + z 2 yb + 2yz zb 
 2 E where, k is a constant with appropriate units. ⃗ × E⃗ ̸= 0, then it is not a valid electrostatic field. If ∇ ⃗ × E⃗ = 0, then it is a valid electrostatic field. If ∇ Applied Physics August 3, 2024 37 / 66 Example 7 – (2) xb yb zb ⃗ × E⃗1 = k ∇ ∂ ∂ ∂ x (0 − 2y ) + yb (0 − 3z) + zb (0 − x)] = k [b ∂x ∂y ∂z xy 2yz 3zx = k (−2y xb − 3z yb − x zb) ̸= 0 xb yb zb ⃗ × E⃗2 = k ∇ ∂ ∂ ∂ ∂x ∂y ∂z y2 2xy + z 2 2yz x (2z − 2z) + yb (0 − 0) + zb (2y − 2y )] = k [b =k ·0=0 Applied Physics August 3, 2024 38 / 66 Electric Potential – (1) The electric field E⃗ is not just any old vector function. It is a very special kind of vector function: one whose curl is zero. For example: E⃗ = y xb cannot possibly be an electrostatic field. If curl of any vector, E⃗ is zero, it can be written as gradient of a scalar potential V. ∇⃗ × E⃗ = 0 ⇔ E⃗ = −∇V ⃗ ⃗ × E⃗ = 0, the line integral of E⃗ around any closed loop is zero. Because ∇ Applied Physics August 3, 2024 39 / 66 Electric Potential – (2) Because E⃗ · d ⃗l = 0, the line integral of E⃗ from point a to point b is the H same for all paths. Because the line integral is independent of path, a function V (⃗r ) can be defined as, Z ⃗r V (⃗r ) = − E⃗ · d ⃗l O Here, O is some standard reference point. V depends only on ⃗r. It is called electric potential. Applied Physics August 3, 2024 40 / 66 Electric Potential – (3) The potential difference between two points a and b is, Z b Z a V (b) − V (a) = − E⃗ · d ⃗l + E⃗ · d ⃗l O O Z b Z O =− E⃗ · d ⃗l + E⃗ · d ⃗l O a Z b =− E⃗ · d ⃗l a Fundamental theorem of gradients Z b   V (b) − V (a) = ⃗ ∇V · d ⃗l a Applied Physics August 3, 2024 41 / 66 Electric Potential – (4) So...... Z b Z b   − E⃗ · d ⃗l = ⃗ ∇V · d ⃗l a a Since, this is true for any points a and b, the integrands must be equal: E⃗ = −∇V ⃗ Important to note: Notice the subtle but crucial role played by path independence in this argument. If the line integral of E⃗ depended on the path taken, then the “definition” of V would be nonsense. Applied Physics August 3, 2024 42 / 66 Comments on potential – (1) (i) The Name The word “potential” is a hideous misnomer because it inevitably reminds you of potential energy. This is particularly insidious, because there is a connection between “potential” and “potential energy”. It should be emphasised once and for all that “potential” and “potential energy” are completely different terms and should, by all rights, have different names. Incidentally, a surface over which the potential is constant is called an equipotential. Applied Physics August 3, 2024 43 / 66 Comments on potential – (2) (ii) Advantage of the potential formulation If V is known, E⃗ can be easily calculated by taking the gradient: E⃗ = −∇V⃗. Since E⃗ is a vector quantity (three components), but V is a scalar (one component), a genuine question can be: How can one function possibly contain all the information that three independent functions carry? The answer is that the three components of E are not really as independent as they look; in fact, they are explicitly interrelated by the ⃗ × E⃗ = 0. very condition, ∇ Applied Physics August 3, 2024 44 / 66 Comments on potential – (3) (iii) The reference point O There is an essential ambiguity in the definition of potential, since the choice of reference point O was arbitrary. Changing reference points amounts to adding a constant K to the potential. Z ⃗ r Z O Z ⃗ r ′ V (⃗r ) = − E⃗ · d ⃗l = − E⃗ · d ⃗l − − E⃗ · d ⃗l = K + V (⃗r ) O′ O′ O Here K is the line integral of E⃗ from the old reference point O to the new one O ′. Applied Physics August 3, 2024 45 / 66 Comments on potential – (4) (iv) Potential obeys the superposition principle The original superposition principle pertains to the force on a test charge Q. It says that the total force on Q is the vector sum of the forces attributable to the source charges individually. Dividing through by Q, it can be shown that the electric field, too, obeys the superposition principle. Integrating from the common reference point to ⃗r , it follows that the potential also satisfies such a principle, V = V1 + V2 + V3 +... Applied Physics August 3, 2024 46 / 66 Comments on potential – (5) (v) Units of Potential In SI units, force is measured in newtons and charge in coulombs. So, the electric field is measured in the units of Newtons per coulomb. Accordingly, potential is newton-meters per coulomb, or joules per coulomb. A joule per coulomb is a volt. Applied Physics August 3, 2024 47 / 66 Example 8 – (1) Find the potential inside and outside a spherical shell of radius R that carries a uniform surface charge. Set the reference point at infinity. In such cases, the electric field is obtained from the Gauss’s Law. Field outside Field inside 1 q E⃗ = rb E⃗ = 0 4πϵ0 r 2 Applied Physics August 3, 2024 48 / 66 Example 8 – (2) Potential outside the spherical surface, In this situation, r > R, the path integral, d ⃗l = dr rb. Limits of integral, r : ∞ → r. Z P Z r 1 q ′ q V (⃗r ) = − E⃗ · d ⃗l = − dr = ∞ 4πϵ0 ∞ r ′2 4πϵ0 r Potential inside the spherical surface To find the potential inside the sphere (r < R), the integral is split into two pieces, one for ∞ → R and another for R → r. Z R Z r Z R q V (⃗r ) = − E⃗ · d ⃗l − E⃗ · d ⃗l = − dr − 0 ∞ R ∞ r2 q = 4πϵ0 R Applied Physics August 3, 2024 49 / 66 Work it takes to move a charge – (1) Suppose there is a stationary configuration of source charges, and a test charge Q from point ⃗a to point ⃗b. Question What is the work done to move the charge? At any point along the path, the electric force on Q is F⃗ = Q E⃗. Hence, the force required in opposition to this electrical force, is −Q E⃗. The work done is therefore, Z ⃗ b Z ⃗ b h   i W = F⃗ · d ⃗l = −Q E⃗ · d ⃗l = Q V ⃗b − V (⃗a) ⃗ a ⃗ a Applied Physics August 3, 2024 50 / 66 Work it takes to move a charge – (2) The work done is independent of the path taken from ⃗a to ⃗b. In mechanics terms, the electric force is conservative. Dividing through by Q, V (b) − V (a) = W /Q In words, The potential difference between points ⃗a and ⃗b is equal to the work per unit charge required to carry a particle from ⃗a to ⃗b. If Q is brought from a point faraway to a point ⃗r then the work In this sense, potential is potential done is, energy (the work it takes to create the system) per unit charge. W = Q [V (⃗r ) − V (∞)] = QV (⃗r ) Applied Physics August 3, 2024 51 / 66 Energy of a point charge distribution – (1) How much work would it take to assemble an entire collection of point charges? The first charge, q1 , takes no work, since there is no field yet to fight against. To bring charge q2 to a point r⃗2 , To bring charge q3 to a point r⃗3 , the work done is q2 V1 (⃗ r1 ), V1 is the work done is potential due to q1. q3 V1 (⃗ r3 ) + q3 V2 (⃗ r3 ). q2 q1   q3 q1 q2 W2 = W3 = + 4πϵ0 r 12 4πϵ0 r 13 r 23 Applied Physics August 3, 2024 52 / 66 Energy of a point charge distribution – (2) To bring charge q4 to a point r⃗4 , the work done is q4 V1 (⃗ r4 ) + q4 V2 (⃗ r4 ) + q4 V3 (⃗ r4 ).   q4 q1 q2 q3 W4 = + + 4πϵ0 r 14 r 24 r 34 The total work necessary to assemble the first four charges, then, is   1 q1 q2 q1 q3 q1 q4 q2 q3 q2 q4 q3 q4 W = + + + + + 4πϵ0 r 12 r 13 r 14 r 23 r 24 r 34 Applied Physics August 3, 2024 53 / 66 Energy of a point charge distribution – (3) In summation terms for n charges, Separating summation of qi and rearranging, n n 1 X X qi qj W =   4πϵ0 r ij i=1 j>i n n 1 X  X 1 qj  W = qi 2 4πϵ0 r ij i=1 j̸=i j > i implies that the same pair is not counted twice. Although, it is The term in parentheses is, total easier to divide by 2 after counting potential at point ⃗ri due to the the pair twice. other charges qj. n n n 1 1 X X qi qj 1X W = 2 4πϵ0 r ij i=1 j̸=i ∴W = 2 qi V (⃗ri ) i=1 Applied Physics August 3, 2024 54 / 66 Energy of a continuous charge distribution – (1) For a volume charge density, Substituting ρ in work done, The work done to assemble the Z ϵ0  ⃗ ⃗  system is, W = ∇ · E Vdτ 2 Z 1 W = ρVdτ Using integration by parts, 2 Z ϵ0   R For line charge, λVdl. W =− E⃗ · ∇V⃗ dτ R 2 For surface charge, σVda. ϵ0 Z   + ∇⃗ · V E⃗ dτ 2 Z From Gauss’s law, the charge ϵ0 W = E 2 dτ density is, 2 V I ϵ0 ⃗ · E⃗ ρ = ϵ0 ∇ + V E⃗ · d⃗a 2 S Applied Physics August 3, 2024 55 / 66 Energy of a continuous charge distribution – (2) What should be the volume of integration? From its derivation, it is clear that the integration should be over the region where the charge is located. But actually, any larger volume would do just as well. The “extra” territory will contribute nothing to the integral, since ρ = 0 out there. By enlarging the volume beyond the minimum necessary to trap all the charges, the E 2 integral can only increase. Since, work done is to be constant, the surface integral must decrease correspondingly. If the integral is taken all over the space, the surface integral goes to zero. Z ϵ0 ∴W = E 2 dτ 2 Applied Physics August 3, 2024 56 / 66 Example 9 – (1) 1 Three charges are situated at the comers of a square (side a). How much work does it take to bring in another charge, +q, from far away and place it in the fourth corner? 2 How much work does it take to assemble the whole configuration of four charges? Work done for 1st charge (top left corner) W1 = 0. Total work done, W = W1 + W2 + W3 + W4. Applied Physics August 3, 2024 57 / 66 Example 9 – (2) Potential by the three charges, Work done for 2nd charge (bottom left corner), 1 X qi V = 4πϵ0 r ij −q 2 W2 = 4πϵ0 a   1 q 2q V = √ − 4πϵ0 2a a Work done for 3rd charge (bottom right corner), Work done for 4th charge, q2 q2 W3 = √ − q2   1 4πϵ0 2a 4πϵ0 a W4 = qV = √ −2 4πϵ0 a 2 Applied Physics August 3, 2024 58 / 66 Problems for Homework – (1) Problem 1 Seven equal charges, q, are situated at the top corners of a regular 12-sided polygon.Calculate the net force on a test charge Q at the center. Problem 2 If 13 equal charges, q, are placed at the corners of a regular 13-sided polygon. Calculate the force on a test charge Q at the center. Problem 3 Suppose one of the 13 q’s is removed. What is the force on Q? Explain your reasoning carefully. Applied Physics August 3, 2024 59 / 66 Problems for Homework – (2) Problem 4 Find the electric field a distance z above the midpoint between two equal but opposite charges (q and −q), a distance d apart. Problem 5 Find the electric field a distance z above one end of a straight line segment of length L that carries a uniform line charge λ. Check that your formula is consistent with what you would expect for the case z >> L. Applied Physics August 3, 2024 60 / 66 Problems for Homework – (3) Problem 6 Find the electric field a distance z above the center of a square loop (side a) carrying uniform line charge λ. Problem 7 Find the electric field a distance z above the center of a circular loop of radius r that carries a uniform line charge λ. Applied Physics August 3, 2024 61 / 66 Problems for Homework – (4) Problem 8 Find the electric field a distance z above the center of a flat circular disk of radius R that carries a uniform surface charge σ. What does your formula give in the limit R → ∞? Also check the case z >> R. Problem 9 Suppose the electric field in some region is found to be E⃗ = kr 3 rb, in spherical coordinates (k is some constant). 1 Find the charge density ρ. 2 Find the total charge contained in a sphere of radius R, centered at the origin. Applied Physics August 3, 2024 62 / 66 Problems for Homework – (5) Problem 10 A charge q sits at the back comer of a cube. What is the flux of E⃗ through the shaded side? Problem 11 Two infinite parallel planes carry equal but opposite uniform charge densities ±σ. Find the field in each of the three regions: (i) to the left of both, (ii) between them, (iii) to the right of both. Applied Physics August 3, 2024 63 / 66 Problems for Homework – (6) Problem 12 Use Gauss’s law to find the electric field inside and outside a spherical shell of radius R that carries a uniform surface charge density σ. Problem 13 Use Gauss’s law to find the electric field inside a uniformly charged solid sphere (charge density ρ). Problem 14 Find the electric field a distance s from an infinitely long straight wire that carries a uniform line charge λ. Applied Physics August 3, 2024 64 / 66 Problems for Homework – (7) Problem 15 Find the electric field inside a sphere that carries a charge density proportional to the distance from the origin, ρ = kr , for some constant k. Problem 16 Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of V in each region, and check that it yields the correct field. Problem 17 Find the potential a distance s from an infinitely long straight wire that carries a uniform line charge A.. Compute the gradient of your potential, and check that it yields the correct field. Applied Physics August 3, 2024 65 / 66 Problems for Homework – (8) Problem 18 Two positive point charges, qA and qB (masses mA and mB ) are at rest, held together by a massless string of length a. Now the string is cut, and the particles fly off in opposite directions. How fast is each one going, when they are far apart? Applied Physics August 3, 2024 66 / 66