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Chapter 2 Action at a distance 2.1 The : 2.1.1 Introduction What is the force on the test charge Q due to a source ’s law, like Newton’s law of gravitation, involves the charge q? concept of action at a distance. We shall consider the special case of the electrostatics in It simply states how the particles interact but provides no which all the source charges are stationary. explanation of the mechanism by which the force is The principle of superposition states that the interaction transmitted from one point to the other. between any two charges is completely unaffected by the Even Newton himself is not comfortable with this aspect of presence of others. his theory.

What is the concept of action at a distance? This leads to the gravitational, electric, and .

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2.1.2 Coulomb’s Law 2.1.3 The Electric Field How does one particle sense the presence of the other? Coulomb’s law quantitatively describe the interaction of charges. The creates an electric field in the space around it. A second charged particle does not interact Coulomb determined the force law for electrostatic charges directly with the first; rather, it responds to whatever field it directly by experiment. encounters. In this sense, the field acts as an intermediary πε kqQ 1 qQ between the particles. F = 2 rˆ = 2 rˆ Q q1 q2 r 4 0 r F = F + F +πε" = ( ˆ + ˆ +") = QE πε 1 2 2 r1 2 r2 2 4 0 r1 r2 9 N ⋅m n Where k = 9×10 2 1 q q 1 q C 1 2 " i where E = ( 2 rˆ1 + 2 rˆ2 + ) = ∑ 2 rˆi 2 4 0 r1 r2 4πε 0 i=1 ri −12 C and ε 0 = 8.86×10 N ⋅m2 The electric field strength is defined as the force per

3 unit charge placed at that point. 4 Example 2.1.4 Continuous Charge Distributions On a clear day there is an electric field of approximately In order to find the electric field due to a continuous distribution 100 N/C directed vertically down at the earth’s surface. of charge, one must divide the charge distribution into Compare the electrical and gravitational forces on an elements of chargeε dq which may be considered to . be point charges. 1 dq 1 dq dE = ˆ ⇒ E = ˆ Solution: 2 r ∫ 2 r 4π 0 r 4πε 0 r The magnitude of the electrical force is Fe=eE=1.6x10-19x100=1.6x10-17 N. (upward)

The magnitude of the gravitational force is Fg=mg=9.11x10-31x9.8=8.9x10-30 N. (downward)

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Example 2.1 Example Non-conducting disk of radius a has a uniform surface What is the field strength at a distance R from an infinite σ C/m2. What is the field strength at a line of charge with linear charge density λ C/m. distance y from the center along the central axis. Solution: Solution: Since the is infinite long, the electric field in y-direction completely The y-component of the field is cancelλ out. Thus the resultant field is θ kdq y alongπε the x-axis. dEy == dE cosθ rr2 λ 2 222 1λ cos dAπε 1 θRsec cos d where rdqxdx=+x y and = (2 ) dE = θ = x πε4 rθ2 4 (Rθsecθ )2 πσa 2xdx 0 0 Eky= θ y ∫0 223/2 σπ 1 cos d ()xy+ = λ 2 4πε R a dx 0 = kyπσ θ ∫0 223/2 1 π / 2 θ λ ()xy+ Ex = cos d = ∫− / 2 4 0 R π 2πε0 R 7 8 2.2 Divergence and Curl of Electrostatics Fields Field Lines 2.2.1 Field Lines, , and Gauss’s Law How to determine the field strength from the field lines? How to express the magnitude and vector properties of the field strength? The lines are crowed together when the field is strong and spread apart where the field is weaker. The field strength is The field strength at any point could be represented by an proportional to the density of the lines. arrow drawn to scale. However, when several charges are present, the use of arrows of varying and orientations becomes confusing. Instead we represent the electric field by continuous field lines or lines of force.

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Example Flux

Sketch the field lines for two point charges 2Q and –Q. The ΦE through this surface is defined as

Solution: Φ E = EAcosθ = E⋅ A (a)Symmetry (b)Near field (c)Far field For a nonuniform electric field (d)Null point (e)Number of lines Φ = E⋅nˆda E ∫

11 12 Flux Gauss’s Law Flux leaving a closed surface is positive, whereas flux How much is the flux for a spherical around entering a closed surface is negative. a point charge? The net flux through the surface is zero if the number of The total flux through this closed lines that enter the surface is equal to the number that Gaussian surface is leave. Φ = E⋅nˆda E ∫ πε π Q 2 Q = 2 ⋅4 r = 4 0r ε 0

The net flux through a closed surface equals 1/ε0 times the net charge enclosed by the surface.

Can we prove the above statement for arbitrary closed shape? 13 14

Gauss’s Law (II) Turn Gauss’s Law from integral equation into differential form •The circle on the integral sign indicates that the Gaussian Q surface must be enclosed. E⋅da = E⋅nˆda = enc ∫S ∫S •The flux through a surface is determined by the net charge ε 0 enclosed. By applying the ε Qenc 1 ρ E⋅nˆda = (∇ ⋅E)dτ and = ε dτ ∫S ∫v ∫v τ 0 0 1 ρ So (∇ ⋅E)d = ε dτ ∫v ∫v 0 Since this holds for any , the integrands must be equal How to apply Gauss’s law? 1 ∇ ⋅E = ρ Gauss’s Law in differential form. 1. Use symmetry. ε 0 2. Properly choose a Gaussian surface (E//A or E⊥A). 15 16 2.2.2ε The Divergence of Electric Field & Example 2.2 2.2.3 Application of the Gauss’s Law The electric field can be expressed in the following form A non-conducting uniform charged sphere of radius R has ε a total charge Q uniformly distributed throughout its volume. 1 dq 1 rˆ ρ E = rˆ = (r′)dτ ′ Find the field (a) inside, and (b) outside, the sphere. ε ∫all space 2 ∫all space 2 4π 0 r 4π 0 r ρ Solution: Divergence of the electric fieldτ is (a) inside Why the divergence πδ 4 3 1 ˆ Φenc Q 3 r 1 r ′ ′ operator doesn’t apply E = rˆ = ( ) rˆ ∇ ⋅E = (∇ ⋅ 2 ) (r )d π 2 π4 3 2 ∫all space 4 r R 4 0r 4π 0 r on the r’ coordinate? 3 ε Q π rˆ πδ3 = rrˆ πε Since (∇ ⋅ ) = 4 ( ), 3 2 r 4πε R r ρ 0 1 τ 1 (b) outside ∇ ⋅E = 4 3 (r − r′) (r′)d ′ = ρ(r) ∫all space ε πΦ Q 4π 0 0 ˆ ˆ E = 2 r = 2 r 4 r 4πε0r 17 18

Example 2.3 Example 2.5 Find the field due to the following: (a) an infinite sheet of A long cylinder carries a charge density that is proportional charge with density +σ; (b) two parallel to the distance from the axis: λ=ks, for some constant k. infinite sheets with charges density +σ and -σ. Find the electric field inside the cylinder? Solution: Solution: Pick up a Gaussian surface as shown in the figure. The total charge enclosed is

s 2 3 Qenc = A (ks′)s′ds′d = kAs ∫0 φ 3 π Qenc 1 2 E = ε = ks in sˆ direction 0 2πsA 3

19 20 ε ρ How to Choose a Good Gaussian Surface? 2.2.4 The Curlτ of the Electric Field The electric field can be expressedε in the following form Gauss’s Law is always true, but it is not always useful. 1 ˆ 1 1 ρ Symmetry is crucial to the application of Gauss’s law. E = r (r′)d ′ = (∇ ) (r′)dτ ′ 4π ∫all space 2 4π ∫all space There are only three kinds of symmetry that work: 0 r 0 r ε Curl of the electric field is Why the divergence 1. Spherical symmetry: Make your Gaussian surface a 1 1 ρ operator doesn’t apply concentric sphere. ∇×E = (∇×(∇ )) (r′)dτ ′ ∫all space on the r’ coordinate? 2. Cylindrical symmetry: Make your Gaussian surface a 4π 0 r coaxial cylinder. Curl of is always zero. ∴∇×E = 0 3. Plane symmetry: Use a Gaussian “pillbox”, which straddles the surface. The principle of superposition states that the total field is

a vector sum of their individual fields E=E1+E2+… ∇×E = ∇×(E + E +") = ∇×E + ∇×E +" = 0 21 1 2 1 2 22

2.3 Electrical Potential Mechanical Analogous of Potential 2.3.1&2 Introduction to and Comments on Potential The motion of a particle with positive charge q in a uniform electric field is analogous to the motion of a particle of mass Can we apply the concept of potential, first introduced in m in uniform gravitational field near the earth. mechanics, to electrostatic system and find the law of conservation of energy? WEXT = +∆U = U f −Ui We can define an electrostatic potential energy, analogous to gravitational potential energy, and apply the law of If WEXT >0, work is done by the conservation of energy in the analysis of electrical problems. external agent on the charges. If WEXT <0, work is done on the Potential is a property of a point in space and depends only external agent by the field. on the source charges.

Potential energy depends not only on the “source” but also Potential is not equal to the potential energy. on the “test” particle. Thus it will be more convenient if we can define a potential function which is function of “source” 23 only. 24 The Unit of Potential: Volt Only Changes in Potential are Significant When a charge q moves between two points in the We see that only changes in potential ∆V, rather than the electrostatic field, the change in , ∆V, is specific value of Vi and Vf, are significant. defined as the change in electrostatic potential energy per It is convenient to choose the ground connection to earth as unit charge, ∆U ∆V = the zero of potential. q The potential at a point is the external work need to bring The SI unit of electric potential is the volt (V). a positive unit charge, at constant speed, from the position 1 V =1 J / C =1 N ⋅m / C of zero potential to the given point. In an external electric field, both positive and negative The quantity V depends only on the field set up by the charges tend to decrease the electrostatic potential energy. source charges, not on the test charge. Which side will a charge particle drift if it is in the middle of two conducting plates with potential difference, higher or WEXT = q∆V = q(V f −Vi ) lower potential side? 25 26

Potential is Conservative Differential form of Potential In mechanics, the definition of potential energy in terms of The fundamental theorem of gradient states that the work done by the conservative force is ∆U=-Wc.. The B negative sign tells us that positive work by the conservative VB −VA = (∇V )⋅ds ∫A force leads to a decrease in potential energy. B Therefore, the change in potential energy, associated with and V B − V A = − E ⋅ d s so E = −∇V an infinitesimal displacement ds, is ∫A

dU = −Fc ⋅ds = −qE⋅ds The electric field E is a very special kind of vector function dU whose curl is always zero. dV = = −E⋅ds q ∇×E = −(∇×∇V ) = 0 B VB −VA = − E⋅ds ∫A It is often easier to analyze a physical situation in terms of Since the electrostatic field is conservative, the value of potential, which is a , rather than the electric field this depends only on the end points A and B, strength, which is a vector. not on the path taken. 27 28 Example 2.6 Find the potential inside and outside a 2.3.3 Poisson’s Equation and Laplace’s Equation spherical shell of radius R, which carries a uniform surface charge. Set the reference point at infinity. The electric field can be written as the gradient of a scalar potential. Sol : Use the Gauss's law to find the electric field E = −∇V and then use the electric field to calculate the potential. What do the fundamental equations for E looks like, in terms of V? Inside (r < R) E = 0  2 ρ  q Gauss's law ∇ ⋅E = −(∇⋅∇V ) = −∇ V = outside (r > R) E =  2 ε 0  4π 0r ε Curl law ∇×E = −(∇×∇V ) = 0 r q V (r) = − E⋅dA = (r > R) ∫∞ 4π r ε 0 q ∇×E = 0 permits E = −∇V; and V (r) = (r < R) 4πε 0 R in turn, E = −∇V guarantees ∇×E = 0 29 30

2.3.4 The Potential of a Localized Charge The Potential of a Localized Charge Distribution Distribution Setting the reference point at infinity, the potential of a point In general, the potential of a collection of charges is charge q at the origin is 1 n q ε V (r) = ∑ i −1 r q 1 q 4πε V (r) = dr′ = 0 i=1 ri 4π ∫∞ r′2 4πε r 0 0 For a continuous distribution 1 dq The conventional minus sign in the definition of V was V (r) = ∫ chosen precisely in order to make the potential of a positive 4πε 0 r charge come out positive. For a volume charge ρ, a surface charge σ, a line charge λ. 1 ρ(r′) 1 σ (r′) 1 λ(r′) V (r) = ∫ dτ ′ ∫ dτ ′ ∫ dτ ′ 4πε 0 r 4πε 0 r 4πε 0 r

31 32 Example Example A nonconducting disk of radius a has a uniform surface A shell of radius R has a charge Q uniformly distributed charge density σ C/m2. What is the potential at a point on over its surface. Find the potential at a distance r >R from theπε axis of the disk at a distance from its center. its center. σ π Solution:σπ Solution: dq dVπε= , dq = (2 xdx) It is more straightforward to use the electric field, which we 4 0r know from Gauss’s law.

σπ 2 r dV = dx Q r Q 1  1 2 2 ˆ V (r) −V (∞) = − dr = − Q − πε E = 2 r ∫∞ 2   4 0 x + y 4 0r 4 0  r 0 4πε0r σ a Q πε V = dx2 V (r) = ∫0ε 2 2 4πε0r πε 4 0 x + y σ 2 2 0.5 a 2 2 0.5 The potential has a fixed value at all points within the = [](x + y ) − y 0 = [](a + y ) − y 2 0 2ε 0 conducting sphere equal to the potential at the surface. 33 34

2.3.5 Summary; Electrostatic Boundary Conditions Electrostatic Boundary Conditions: Normal

The electric field is not We have derived six formulas continuous at a surface with interrelating three fundamental charge density σ. Why? quantities: ρ , E and V. Consider a Gaussian pillbox. Gauss’s law states that Q σA E⋅da = enc = ∫S ε ε These equations are obtained from two observations: 0 0 The sides of the pillbox contribute nothing to the flux, in the •Coulomb’s law: the fundamental law of electrostatics limit as the thickness ε goes to zero. •The principle of superposition: a general rule applying to ⊥ ⊥ σA ⊥ ⊥ σ all electromagnetic forces. (Eabove − Ebelow )A = ⇒ (Eabove − Ebelow ) = ε 0 ε 0

35 36 Electrostatic Boundary Conditions: Tangential Boundary Conditions in terms of potential σ

The tangential component of E, ε ˆ ˆ by contract, is always continuous. Eabove − Ebelow = n ⇒ (∇Vabove − ∇Vbelow ) = − n 0 0 σ ∂V ∂V σ Consider a thin rectangular loop. or ( above − below ) = − E⋅dA = 0 ∂n ∂n ε ε The curl of the electric field states that ∫P 0 ∂V where above = ∇V ⋅nˆ The ends gives nothing (as εÆ0), and the sides give ∂n // // // // denotes the normal derivative of V. (Eabove − Ebelow )A = 0 ⇒ Eabove = Ebelow

σ Why? In short, Eabove − Ebelow = nˆ, and Vabove = Vbelow ε 0 37 38

2.4 Work and Energy in Electrostatics 2.4.1 The Work Done to Move a charge Homework #3 How much work will you have to do, if you move a test charge Q from point Problems: 2.9, 2.12, 2.15, 2.20, 2.25 a to point b?

What we’re interested is the minimum force you must exert to do the job. b b W = − F ⋅dA = −Q E⋅dA = Q(V (b) −V (a)) ∫a ∫a So V (b) −V (a) = W / Q 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. 39 40 2.4.2 The Energy of a Point Charge Distribution Potential and Potential Energy: Motion of Charges

How much work would it take to The motion of a charge in an electric field may be discussed assemble an entire collection of point in terms of the conservation of energy, ∆K+∆U =0. In terms of charges?πε potential, the conservation law may be written as πε 1 q 1 q q W = 0, W = q ( 1 ) , W = q ( 1 + 2 ) ∆K = −q∆V 1 2 4 2 πε 3 4 3 0 r12 0 r13 r23 It is convenient to measure the energy of elementary 1 q q q q q q W = ( 1 2 + 1 3 + 2 3 ) particles, such as and protons, in terms of a non-SI unit called the electronvolt (1 eV=1.6x10-19 J). 4 0 r12 r13 r23 n n n n πε 1 qiq j 1 qiq j The general rule : W = ∑∑ = ∑∑ According to Einstein famous E=mc2, find the energy in terms 4 0 i=11j= rij 8 0 i=11j= rij -31 j>i j≠i of eV for an electron of rest mass 9.1x10 kg, where the πε 8 n n n speed of light is 3x10 m/s. 1 1 q j 1 = ∑∑qi ( ) = ∑ qiVi (ri ) E=9.1x10-31x(3x108)2/1.6x10-19=0.511 MeV 2 i=114πε 0 j= rij 2 i=1 j≠i 41 42

Example 2.4.3 The Energy of a Continuousρ Charge Distribution ρ A proton, of mass 1.67x10-27 kg, enters the region between τ τ Generalizing the point charge distribution result: parallel plates a distance 20 cm apart. There is a uniform ε 1 1 electric field of 3x105 V/m between the plates, as shown below. dW = (dq )V (r ) = V (r )(d )ε i 2 i i i 2 i i i τ If the initial speed of the proton is 5x106 m/s, what is its final speed? 1 1 0 W = Vd = ( 0∇ ⋅E)Vd = (∇ ⋅E)Vdτ 2 ∫ 2 ∫ 2 ∫ Solution:

1 1 : ∇ ⋅(VE) = (∇V )⋅E + (∇ ⋅E)V mv2 − mv2 = −qV 2 f 2 i ε 0 ε 0 W = (∇ ⋅E)Vdτ = []− E⋅(∇V )d + (EV )da v = v2 + 2qV / m ε2 ∫ 2 ∫ ∫ f i = ((5×106 )2 − 2×1.6×10−19 ×−6×104 /1.67×10−27 ))0.5 0 2 = []E dτ + (EV )da 6 2 ∫ ∫S τ divergence theorem = 6×10 m/s. ε W = 0 E 2dτ ∫ 43 44 2 all space Potential and Potential Energy of Point Charges Example In 1913, Bohr proposed a model of the hydrogen in which an electron orbits a stationary proton in a circular path. Find the total mechanical energy of the electron given that the radius of the orbit is 0.53x10-10 m. Solution: The mechanical energy is the sum of the kinetic and potentialπε energies, E=K+U. The centripetal force is provided by the coulomb attraction. e2 U = −πε 4 0r 2 2 2 e mv 1 2 e F = 2 = ⇒ K = mv = 4 0r r 2 8πε0r 1 9×109 ×(1.6×10−19 )2 E = U + K = − U = − = −2.18×10−18 J = −13.6 eV 45 2 2×0.53×10−10 46

Example 2.4.4 Comments on Electrostatic Energy A metal sphere of radius R has a charge Q. Find its (i) A perplexing “inconsistency” potential energy. ε Solution: W = 0 ∫ E 2dτ ≥ 0 2 all space Which equation is correct? q 1 n dW = Vdq =πε dq W = q V (r ) ≥ or ≤ 0 Both equations are correct. 4 r ∑ i i i 0 2 i=1 2 Q πεq Q W = dq = ∫0 4 0r 8πε0 R Why the energy of a point charge is infinite? Does it make sense? No πε The potential energy U=1/2QV is θ ε 0 2 ε 0 q 2 2 θ the work needed to bring the W = E dτ = ( ) (r sin drd dφ) = ∞ 2 ∫ 2 ∫ 4 r 2 system of charges together. all space 0

47 48 Comments on Electrostatic Energy 2.5 Conductor 2.5.1 Basic Properties (ii) Where is the energy stored? ε 1 n W = ( 0 E 2 )dτ W = q V (r ) E = 0 inside a conductor ∫ ∑ i i i all space 2 2 i=1 It is unnecessary to worry about where the energy ρ = 0 inside a conductor is located. Really? (iii) Superposition principle is not valid, because the Any net charge resides on the surface electrostatic energy is quadratic in the fields. ε ε ε W = 0 E 2dτ = 0 (E + E )2 d A conductor is an equipotential ∫ ∫ 1 2 τ 2 all space 2 all space

= 0 (E 2 + E 2 + 2E ⋅E )dτ E is perpendicular to the surface, just outside a conductor. ∫ 1 2 1 2 2 all space 49 50

Charge Redistribution Discharge at Sharp Points on a Conductor

Suppose two charged metal spheres with radius R1 and R2 are connected by a long wire. Charge will flow from one to σ 1 the other until their potential are equal. The equality of the E = ∝ potential implies that ε 0 R

Q Q 1 = 2 , since Q = 4 R2 The above equation infer that the field strength is greatest at R R the sharp points on a conductor. σ 1 2 π 6 1R1 = σ 2 R2 σ If the field strength is great enough (about 3x10 V/m for dry air) it can cause an electrical discharge in air. We infer that 1/R: The surface charge density on each sphere is inversely proportional to the radius. How does the breakdown occur in high transmission The regions with the smallest radii of curvature have the line? greatest surface charge densities. 51 52 Dust Causing High Voltage Breakdown 2.5.2 Induced Charges Induced charge on metal sphere The potential at the surface of a charged sphere is V=kQ/R and the field strength is E=kQ/R2. So, for a given breakdown field strength, breakdown voltage is proportional to the radius, VB ∝ R. The potential of a sphere of radius 10 cm may be raised to 3x105 V before breakdown. On the other hand, a 0.05 mm If there is some cavity in the conductor, and within that dust particle can initiate a discharge at 150 V. cavity there is some charge, then the field in the cavity will not be zero.

No external fields penetrate the conductor; they are A high voltage system must keep at very clean condition. canceled at the outer surface by the induced charge there.

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2.5.3 Surface Charge and Force on a Conductor 2.5.4 Using energy density viewpoint The magnitude of the charge Q stored on either plate of a is directly proportional to the potential difference V In the immediate neighborhood of the surface, the between the plates. Therefore, we may write energy is ε ε σ Q = CV dW = ( 0 E 2 )dτ = ( 0 ( )2 )dadx = fdadx 2 2 0 Where C is a constant of proportionality σ 2 called the of the capacitor. f = ← the forceε per unit area 2ε 0 The SI unit of a capacitance is the farad This amounts to an outward electrostatic pressure on (F). 1Farad =1 coulomb/volt the surface, tending to draw the charge into the field, regardless of the signε of . The capacitance of a capacitor depends on the geometry σ 2 of the plates (their size, shape, and relative positions) and 2 P = 0 E = the medium (such as air, paper, or plastic) between them. 2 2ε 0 55 56 Parallel-plate capacitor Example ε What is the capacitance of an isolated sphere of radius R? A common arrangementε found in capacitors consists of two plates. σ Q dQ ε A E = = ⇒ V = Ed = ∴ C = 0 Solution: A ε A d 0 0 0 πε where ε is 8.85x10-12 F/m. Q 0 V = ⇒ C = 4πε0 R 4 0 R

If we assume that earth is a conducting sphere of radius Example 2.10 A parallel-plate capacitor with a plate separation 6370 km, then its capacitance would be 710 uF. of 1 mm has a capacitance of 1 F. What is the area of each plate? Is earth a good capacitor? No. −3 Cd 1×10 8 2 A = = −12 =1.13×10 m 57 58 ε 0 8.85×10

Example Example A spherical capacitor consist of two concentric conducting A cylindrical capacitor consists of a central conductor of spheres, as shown in the figure. The inner sphere, of radius radius a surrounded by a cylindrical shell of radius b, as

R1, has charge +Q, while the outer shell of radius R2, has shown below. Find the capacitance of a length L assuming charge –Q. Find its capacitance. that air is between the plates. Solution: Solution: λL λ Er = ε = R2 Q Q 1 1 0 2πrLπε 2 0r E = 2 ⇒ V = − Edr = ( − ) πε ∫R1 b 4 0r 4 0 R2 R1 b Vr = − Er dr = − ln( ) ∫a λ R R 2 0 a C = 4πε ( 1 2 ) πε 0 R − R Q b 2 1 πε = − ln( ) πε 2 0 L a The capacitance happens to be negative quantity. 2πε L C = − 0 Why we are interested only in its magnitude? ln(b / a) Again, we are interested only in the magnitude of the 59 capacitance. 60 Energy Stored in a Capacitor

The energy stored in a capacitor is equal to the work done--- Homework #4 for example, by a battery---to charge it. The work needed to transfer an infinitesimal charge dq from the negative plate to the positive plate is dW=Vdq=q/Cdq. Problems: 2.34, 2.36, 2.39, 2.46, 2.48 The total work done to transfer charge Q is

2 2 Q q Q CV W = dq = = ∫0 C 2C 2

What kind of the potential energy does this work convert? Electric potential energy.

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