Electric Potential Due to Dipole

12 Key excerpts on "Electric Potential Due to Dipole"

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  eBook - ePub

  Electric Field Analysis

  • Sivaji Chakravorti(Author)
  • 2017(Publication Date)
  • CRC Press

    (Publisher)

  For field analysis, it is required that a single dipole be characterized by a vector quantity. As depicted in Figure 5.1, let the magnitudes of the charges be + Q and − Q, respectively, and the distance between them is d. The distance vector d → between the two point charges is considered to be directed from the negative charge to the positive charge. Then the dipole moment of the electric dipole is defined as a vector FIGURE 5.1 The dipole moment of an electric dipole. P → = Q d → ⁢ (5.1) The unit of dipole moment is C.m. 5.2.2 Field due to an Electric Dipole The field due to a single electric dipole can be evaluated as the superposition of the field due to two point charges + Q and – Q, as shown in Figure 5.2. Then the electric potential at the point P due to the electric dipole is given by V P = Q 4 π ε 0 r 1 + - Q 4 π ε 0 r 2 = Q 4 π ε 0 (1 r 1 − 1 r 2) ⁢ (5.2) FIGURE 5.2 Field due to an electric dipole. The distance d between the dipole charges is always much smaller than the distance of P from the two charges. Hence, the line segments r 1 and r 2 will be parallel for all practical purposes. Hence, r 1 = r - d 2 cos θ and r 2 = r + d 2 cos θ ⁢ (5.3) where: r = distance from the centre of the electric dipole to P θ = the angle between the distance vectors d → and r → Thus, assuming r ≫ d, electric potential at a distance r from the electric dipole may be written as follows: V P = Q d cos θ 4 π ε 0 r 2 = p → ⋅ u ˆ r 4 π ε 0 r 2 ⁢ (5.4) From the above equation, it may be seen that the field. due to an electric dipole is two dimensional in nature when represented in spherical coordinate system, as the field depends on r and θ coordinates and not on ϕ coordinate. Then electric field intensity at P can be expressed as follows: E → P = - ∇ → V P = - ∂ V P ∂ r u ˆ r - 1 r ∂ V P ∂ θ u ˆ θ = Q d cos θ 2 π ε 0 r 3 u ˆ r + Q d sin θ 4 π ε 0 r 3 u ˆ θ ⁢ (5.5) Thus, the r and θ components of electric field intensity

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  Halliday and Resnick's Principles of Physics

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  24-4 POTENTIAL DUE TO AN ELECTRIC DIPOLE Learning Objectives After reading this module, you should be able to . . . 24.19 Calculate the potential V at any given point due to an electric dipole, in terms of the magnitude p of the dipole moment or the product of the charge separation d and the magnitude q of either charge. 24.20 For an electric dipole, identify the locations of posi- tive potential, negative potential, and zero potential. 24.21 Compare the decrease in potential with increasing dis- tance for a single charged particle and an electric dipole. ● At a distance r from an electric dipole with dipole moment magnitude p = qd, the electric potential of the dipole is V = 1 4πε 0 p cos θ r 2 for r ⪢ d; the angle θ lies between the dipole moment vector and a line extending from the dipole midpoint to the point of measurement. Key Idea z d O θ +q –q r (–) – r (+) r (–) r (+) r P (a) + z d θ r (–) – r (+) r (–) r (+) (b) + +q –q Figure 24-13 (a) Point P is a distance r from the midpoint O of a dipole. The line OP makes an angle θ with the dipole axis. (b) If P is far from the dipole, the lines of lengths r (+) and r (–) are approximately par- allel to the line of length r, and the dashed black line is approximately perpendicular to the line of length r (–) . Potential Due to an Electric Dipole Now let us apply Eq. 24-27 to an electric dipole to find the potential at an arbitrary point P in Fig. 24-13a. At P, the positively charged particle (at distance r (+) ) sets up potential V (+) and the negatively charged particle (at distance r (–) ) sets up potential V (–) . Then the net potential at P is given by Eq. 24-27 as V = ∑ 2 i =1 V i = V (+) + V (−) = 1 4πε 0 ( q r (+) + −q r (−) ) = q 4πε 0 r (−) − r (+) r (−) r (+) .

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  Fundamentals of Physics, Extended

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  697 24-4 POTENTIAL DUE TO AN ELECTRIC DIPOLE 24-4 POTENTIAL DUE TO AN ELECTRIC DIPOLE Learning Objectives After reading this module, you should be able to . . . 24.19 Calculate the potential V at any given point due to an electric dipole, in terms of the magnitude p of the dipole moment or the product of the charge separation d and the magnitude q of either charge. 24.20 For an electric dipole, identify the locations of posi- tive potential, negative potential, and zero potential. 24.21 Compare the decrease in potential with increasing dis- tance for a single charged particle and an electric dipole. ● At a distance r from an electric dipole with dipole moment magnitude p = qd, the electric potential of the dipole is V = 1 4πε 0 p cos θ r 2 for r ⪢ d; the angle θ lies between the dipole moment vector and a line extending from the dipole midpoint to the point of measurement. Key Idea z d O θ +q –q r (–) – r (+) r (–) r (+) r P (a) + z d θ r (–) – r (+) r (–) r (+) (b) + +q –q Figure 24-13 (a) Point P is a distance r from the midpoint O of a dipole. The line OP makes an angle θ with the dipole axis. (b) If P is far from the dipole, the lines of lengths r (+) and r (–) are approximately par- allel to the line of length r, and the dashed black line is approximately perpendicular to the line of length r (–) . Potential Due to an Electric Dipole Now let us apply Eq. 24-27 to an electric dipole to find the potential at an arbitrary point P in Fig. 24-13a. At P, the positively charged particle (at distance r (+) ) sets up potential V (+) and the negatively charged particle (at distance r (–) ) sets up potential V (–) . Then the net potential at P is given by Eq. 24-27 as V = ∑ 2 i =1 V i = V (+) + V (−) = 1 4πε 0 ( q r (+) + −q r (−) ) = q 4πε 0 r (−) − r (+) r (−) r (+) .

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  Halliday's Fundamentals of Physics, 1st Australian & New Zealand Edition

  • David Halliday, Jearl Walker, Patrick Keleher, Paul Lasky, John Long, Judith Dawes, Julius Orwa, Ajay Mahato, Peter Huf, Warren Stannard, Amanda Edgar, Liam Lyons, Dipesh Bhattarai(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  This shift sets up a dipole moment that points in the direction of the field. This dipole moment is said to be induced by the field, and the atom or molecule is then said to be polarised by the field (that is, it has a positive side and a negative side). When the field is removed, the induced dipole moment and the polarisation disappear. 24.5 Potential due to a continuous charge distribution LEARNING OBJECTIVE After reading this module, you should be able to: 24.5.1 for charge that is distributed uniformly along a line or over a surface, fnd the net potential at a given point by splitting the distribution up into charge elements and summing (by integration) the potential due to each one. KEY IDEAS • For a continuous distribution of charge (over an extended object), the potential is found by (1) dividing the distribution into charge elements dq that can be treated as particles and then (2) summing the potential due Pdf_Folio:535 CHAPTER 24 Electric potential 535 to each element by integrating over the full distribution: V = 1 4 0 ∫ dq r = k ∫ dq r . • In order to carry out the integration, dq can be replaced with the product of either a linear charge density  and a length element (such as dx), or a surface charge density  and an area element (such as dx dy). • In some cases where the charge is symmetrically distributed, a two‐dimensional integration can be reduced to a one‐dimensional integration. When a charge distribution q is continuous (as on a uniformly charged thin rod or disc), the number of charged particles is far too many for us to consider them one by one to find the net potential V at a point P. Instead, we must choose a differential element of charge dq, determine the potential dV at P due to dq, and then integrate over the entire charge distribution. Let us again take the point of zero potential to be at infinity. If we treat the charge element dq as a particle, then we can express the potential dV at point P due to dq as dV = 1 4 0 dq r .

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  Understanding Physics and Physical Chemistry Using Formal Graphs

  • Eric Vieil(Author)
  • 2012(Publication Date)
  • CRC Press

    (Publisher)

  In this equation the basic quantities of each pole are opposite and their relation with the charge of dipole is (in adopting the convention of a dipole receptor): Q Q Q = = -1 2 (C7.4) 1 1 –1 –1 + Current Current Current Charge Charge Charge Potential Potential Potential Pole 1 Pole 2 Dipole V 2 V 1 I I 2 I 1 Q 2 Q 1 V Q GRAPH 6.21 168 Understanding Physics and Physical Chemistry Using Formal Graphs Introduction of this double equality C7.4 into the equation of energy variation, Equation C7.2, pro-vides, by using the embedding principle, the relation between pole potentials and dipole potential: V V V = -1 2 (C7.5) The Formal Graph of construction of the dipole from the two poles uses the preceding equations. Constitutive Property : In the frame of the Formal Graph theory, the capacitive relation between the electric potential of the dipole and its charge is written in a very general way with the help of a capacitance operator, as for a pole (see case study A5 “Electric Charges” in Chapter 4): Q = ˆ C V (C7.6) In electrostatics, the capacitive relation is classically considered as a linear relation: Q C lin = V (C7.7) The Formal Graph approach establishes the capacitive relationship, for all energy varieties, on the basis of a theory of influence between entities. The result is that a capacitive relation has an exponential shape for a pole, giving a nonlinear relation for the dipole (a sigmoid shape for a con-servative dipole). This means that the classical linear relation is an asymptote of the exponential function around the origin. For each pole, its activity is (see case study A5 “Electric Charges” in Chapter 4) Q Q Q a i i m i m Qi i + = = exp V V Δ (C7.8) in which Q i m represents the opposite of the minimum charge Q i min . The consequence of the sym-metry between charges is that the maximum of one pole corresponds to the minimum of the other pole, so the relation between limiting quantities for the poles and the dipole is identical for both poles.

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  eBook - ePub

  Molecular Driving Forces

  Statistical Thermodynamics in Biology, Chemistry, Physics, and Nanoscience

  • Ken Dill, Sarina Bromberg(Authors)
  • 2010(Publication Date)
  • Garland Science

    (Publisher)

  Figure 21.8 ). (If you had a more complex system, say a charge of −3.6 at one end and +5.2 at the other end, you would take it to be a dipole of −3.6 at one end and +3.6 at the other end, plus an additional charge of 1.6 at one end; see Chapter 24 . Here we just focus on the dipole component.) Dipoles are oriented in space, indicated by their vector ℓ pointing from −q to +q. The dipole moment is a vector

  Figure 21.8 A dipole.

  μ = q ℓ .

  (21.16)

  Now we calculate the work of orienting an electric dipole.

  EXAMPLE 21.4 The energy of a dipole in an electric field. Figures 21.9 and 21.10 show a dipole with its center fixed in space. The dipole is subject to an orienting force from an electric field E. Compute the work w that the field performs in rotating the dipole from angle 0 (parallel to the field) to an angle θ:

  w = ∫ f ⋅ d ℓ .

  (21.17)

  Figure 21.9 A dipole of length ℓ = 2a orients in an applied electric field E. The force acting on the dipole is f. The component of the force that acts to orient the dipole has magnitude fc = f sin θ.

  Figure 21.10 Displacement due to dipole orientation. The dipole of length 2a changes angle by an amount dθ, so dℓ = adθ.

  The magnitude of the force on a charge q acting in the direction of the field is qE. The force can be decomposed into two vector components: f

  c

  acting in the circumferential direction to rotate the dipole and f

  s

  acting to stretch the dipole.

  We are interested only in f

  c

  . As Figure 21.9 indicates, this has magnitude fc = f sin θ = Eq sin θ. What is the charge displacement dℓ in the circumferential direction? If a point at radial distance a from a pivot point in the middle of the dipole is rotated through an angle dθ, the circumferential displacement is a dθ (see Figure 21.10 ). Because the force acts in the opposite direction of increasing θ

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  Theory of Electric Polarization

  Dielectrics in Static Fields

  • Bozzano G Luisa(Author)
  • 2012(Publication Date)
  • Elsevier Science

    (Publisher)

  (3.9) The potential due to the dipole itself is not included in the potential φ. If one should take the total potential and insert it in eqn. (3.6) or (3.9), one would obtain expressions containing a self-energy term that is infinitely large. Obviously, it is meaningless to speak about the energy of a charge or a dipole in its own field. When it is not clear from the context that an external field is meant, we will use a subscript 0 to denote the external field (£ 0 ), or the corresponding potential (φ 0 ). POLARIZATION AND ENERGY 93 The calculation leading to the expression for the potential energy of a dipole in an external field, eqn. (3.9), can be extended to the case of a multi-pole of the tt-th order with multipole moment Y (n For a quadrupole, consisting of charges +e at r and r + l x + l 2 , —e at r + l x and r + / 2 , we obtain, by developing all potentials in a Taylor series around φ (ν) and taking the limit e oc, Ι γ -> 0, l 2 0 while Q = e (l x l 2 + M i ) (compare eqn. (1.120)) is kept constant: V q = Q :VV0(r). (3.10) For a general multipole with moment Y {n) we find analogously: V yin) = K W ( ^ V ( r ) , (3.11) where as in eqn. (1.151) ^ denotes an η times repeated inner multiplication, and the exponent η of the nabla operator denotes an η times repeated tensor product. For η = 0, 1, 2, 3 we obtain from the general expression (3.11): η = 0 V e = e, (compare (3.6)) η = 1 V m = m ·φ = — m £, (compare (3.9)) η = 2 1/ = Q :VV(/> = -Q : V E , (3.12) π = 3 K M = £ / ; ν ν ν φ = -LTiVVE. (3.13) From this we may conclude, for instance, that in a homogeneous external field £, where VE = 0, the potential energy of a quadrupole and of all higher multipoles is zero. Thus, in a homogeneous field quadrupoles and all higher multipoles will have no direct physical effects. It is also possible to derive an expression for the potential energy of an arbitrary charge distribution in an external field in terms of the multipole moments of this charge distribution.

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  Fundamentals of Physics, Volume 2

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  The electric field vector at any point is tangent to a field line through that point. The density of field lines in any region is proportional to the magnitude of the electric field in that region. Field lines originate on positive charges and terminate on negative charges. Field Due to a Point Charge The magnitude of the elec- tric field E → set up by a point charge q at a distance r from the charge is E = 1 _____ 4πε 0 |q| ___ r 2 . (22.2.2) The direction of E → is away from the point charge if the charge is positive and toward it if the charge is negative. Field Due to an Electric Dipole An electric dipole con- sists of two particles with charges of equal magnitude q but opposite sign, separated by a small distance d. Their elec- tric dipole moment p → has magnitude qd and points from the negative charge to the positive charge. The magnitude of the electric field set up by the dipole at a distant point on the dipole axis (which runs through both charges) is E = 1 _____ 2πε 0 p _____ z 3 , (22.3.5) where z is the distance between the point and the center of the dipole. Review & Summary Field Due to a Continuous Charge Distribution The electric field due to a continuous charge distribution is found by treating charge elements as point charges and then summing, via integration, the electric field vectors produced by all the charge elements to find the net vector. Field Due to a Charged Disk The electric field magni- tude at a point on the central axis through a uniformly charged disk is given by E = σ ____ 2 ε 0 ( 1 − z _________ √ _ z 2 + R 2 ) , (22.5.5) where z is the distance along the axis from the center of the disk, R is the radius of the disk, and σ is the surface charge density. Force on a Point Charge in an Electric Field When a point charge q is placed in an external electric field E → , the elec- trostatic force F → that acts on the point charge is F → = qE → .

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  eBook - ePub

  Physics for Students of Science and Engineering

  • A. L. Stanford, J. M. Tanner(Authors)
  • 2014(Publication Date)
  • Academic Press

    (Publisher)

  2 . Calculate the corresponding value for the radius of the electron.

  Answer: 1.4 × 10−15 m

  13.6 Problem-Solving Summary

  The basic definition of electric potential is expressed in terms of a difference in potential, V B –

  VA ,

  between two points A and

  B: VB – VA

  is equal to the work

  WA→B

  per unit charge necessary to move a particle with charge q from A to B without changing the kinetic energy of the particle. Because the electrostatic field is conservative, the difference in electric potential between A and B is independent of the path along which

  WA→B

  is calculated. Therefore, in solving problems, we may choose the most convenient path for purposes of computing potential differences.

  The definition of potential difference,

  VB – VA = WA→B /q

  , permits us to assign a value to the electric potential at a single point if we have chosen a reference point at which the potential is assigned the value zero. Usually the potential is chosen to be equal to zero at an infinite distance from a given charge. This choice of a zero of potential is appropriate if the charge distribution being considered has a finite extent; otherwise the position at which the potential is chosen to be equal to zero must be specified.

  The electric potential V of a point charge Q at a distance r from the charge is given by V = kQ/r . The potential at point P caused by a collection of point charges is found by first calculating the potential at P caused by each individual charge. Then the electric potential at P is the algebraic sum of each of the potentials caused by the charges in the entire collection. The algebraic sign of the potential computed for each charge Q is positive if Q is positive and negative if Q is negative.

  The electric potential at a point P caused by a continuous distribution of charge of finite extent is calculated by treating each infinitesimal element of charge dQ as if it were a point charge at a distance r from the point P . Then the potential at P

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  eBook - ePub

  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  V assumes that the electric potential is zero at an infinite distance away from the charge. The total electric potential at a given location due to two or more charges is the algebraic sum of the potentials due to each charge.

  (19.6)

  V =

  k q

  r

  The total potential energy of a group of charges is the amount by which the electric potential energy of the group differs from its initial value when the charges are infinitely far apart and far away. It is also equal to the work required to assemble the group, one charge at a time, starting with the charges infinitely far apart and far away.
• 19.4
  Equipotential Surfaces and Their Relation to the Electric Field
  An equipotential surface is a surface on which the electric potential is the same everywhere. The electric force does no work as a charge moves on an equipotential surface, because the force is always perpendicular to the displacement of the charge.
  The electric field created by any group of charges is everywhere perpendicular to the associated equipotential surfaces and points in the direction of decreasing potential. The electric field is related to two equipotential surfaces by Equation 19.7a , where

  ΔV

  is the potential difference between the surfaces and

  Δs

  is the displacement. The term

  ΔV/Δs

  is called the potential gradient.
  (19.7a)

  E = −

  Δ V

  Δ s
• 19.5
  Capacitors and Dielectrics
  A capacitor is a device that stores charge and energy. It consists of two conductors or plates that are near one another, but not touching. The magnitude q of the charge on each plate is given by Equation 19.8 , where V is the magnitude of the potential difference between the plates and C is the capacitance. The SI unit for capacitance is the coulomb per volt (C/V), or farad (F).
  (19.8)

  q = C V
  The insulating material included between the plates of a capacitor is called a dielectric. The dielectric constant κ of the material is defined as shown in Equation 19.9 , where E0 and E