Equipotential Surface

11 Key excerpts on "Equipotential Surface"

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  eBook - PDF

  Cutnell & Johnson Physics, P-eBK

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  An Equipotential Surface is a surface on which the electric potential is the same everywhere. The easiest Equipotential Surfaces to visualise are those that surround an isolated point charge. According to equation 19.6, the potential at a distance r from a point charge q is V = kq/r. Thus, wherever r is the same, the potential is the same, and the Equipotential Surfaces are spherical surfaces centred on the charge. There are an infinite number of such surfaces, one for every value of r, and figure 19.11 illustrates two of them. The larger the distance r, the smaller is the potential of the Equipotential Surface. 520 Physics The net electric force does no work as a charge moves on an Equipotential Surface. This important characteristic arises because when an electric force does work W AB as a charge moves from A to B, the potential changes according to V B − V A = −W AB /q 0 (equation 19.4). Since the potential remains the same everywhere on an Equipotential Surface, V A = V B , and we see that W AB = 0 J. In figure 19.11, for instance, the electric force does no work as a test charge moves along the circular arc ABC, which lies on an Equipotential Surface. In contrast, the electric force does work when a charge moves between Equipotential Surfaces, as from A to D in the picture. The spherical Equipotential Surfaces that surround an isolated point charge illustrate another character- istic of all Equipotential Surfaces. Figure 19.12 shows two of the surfaces around a positive point charge, along with some electric field lines. The electric field lines give the direction of the electric field, and for a positive point charge the electric field is directed radially outwards. Therefore, at each location on an equipotential sphere the electric field is perpendicular to the surface and points outwards in the direction of decreasing potential, as the drawing emphasises.

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  Physics

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

    (Publisher)

  The electric field points in the direction of decreasing potential. 19.4 Equipotential Surfaces and Their Relation to the Electric Field 535 parallel to the surface. This field component would exert an electric force on a test charge placed on the surface. As the charge moved along the surface, work would be done by this component of the electric force. The work, according to Equation 19.4, would cause the potential to change, and, thus, the surface could not be an Equipotential Surface as assumed. The only way out of the dilemma is for the electric field to be perpendicular to the surface, so there is no component of the field parallel to the surface. We have already encountered one Equipotential Surface. In Section 18.8, we found that the direction of the electric field just outside an electrical conductor is perpendicular to the conduc- tor’s surface, when the conductor is at equilibrium under electrostatic conditions. Thus, the sur- face of any conductor is an Equipotential Surface under such conditions. In fact, since the electric field is zero everywhere inside a conductor whose charges are in equilibrium, the entire conduc- tor can be regarded as an equipotential volume. There is a quantitative relation between the electric field and the Equipotential Surfaces. One example that illustrates this relation is the parallel plate capacitor in Figure 19.15. As Section 18.6 discusses, the electric field E → between the metal plates is perpendicular to them and is the same everywhere, ignoring fringe fields at the edges. To be perpendicular to the electric field, the equi- potential surfaces must be planes that are parallel to the capacitor plates, which themselves are Equipotential Surfaces. The potential difference between the plates is given by Equation 19.4 as ΔV = V B − V A = −W AB /q 0 , where A is a point on the positive plate and B is a point on the negative plate.

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  Since the field lines are not simply radial, the Equipotential Surfaces are no longer spherical but, instead, have the shape necessary to be everywhere perpendicular to the field lines. To see why an Equipotential Surface must be perpendicular to the electric field, consider Figure 19.14, which shows a hypothetical situation in which the perpendicular relation does not hold. If E B were not perpendicular to the Equipotential Surface, there would be a component of E B parallel to the surface. This field component would exert an electric force on a test charge placed on the surface. As the charge moved along the surface, work would be done by this component of the electric force. The work, accord- ing to Equation 19.4, would cause the potential to change, and, thus, the surface could not be an Equipotential Surface as assumed. The only way out of the dilemma is for the electric field to be perpendicular to the surface, so there is no component of the field parallel to the surface. We have already encountered one Equipotential Surface. In Section 18.8, we found that the direction of the electric field just outside an electrical conductor is perpendicular to the conductor’s surface, when the conductor is at equilibrium under electrostatic conditions. Thus, the surface of any conductor is an Equipotential Surface under such conditions. In fact, since the electric field is zero everywhere inside a conductor whose charges are in equilibrium, the entire conductor can be regarded as an equipotential volume. There is a quantitative relation between the electric field and the Equipotential Surfaces. One example that illustrates this relation is the parallel plate capacitor in Figure 19.15. As Section 18.6 discusses, the electric field E B between the metal plates is perpendicular to them and is the same everywhere, ignoring fringe fields at the edges.

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  Halliday and Resnick's Principles of Physics

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

    (Publisher)

  Figure 24-4 shows a family of Equipotential Surfaces associated with the elec- tric field due to some distribution of charges. The work done by the electric field on a charged particle as the particle moves from one end to the other of paths 24-2 Equipotential SurfaceS AND THE ELECTRIC FIELD Learning Objectives After reading this module, you should be able to . . . 24.09 Identify an Equipotential Surface and describe how it is related to the direction of the associated electric field. 24.10 Given an electric field as a function of position, cal- culate the change in potential ΔV from an initial point to a final point by choosing a path between the points and integrating the dot product of the field E → and a length element d s → along the path. 24.11 For a uniform electric field, relate the field magni- tude E and the separation Δx and potential difference ΔV between adjacent equipotential lines. 24.12 Given a graph of electric field E versus position along an axis, calculate the change in potential ΔV from an initial point to a final point by graphical integration. 24.13 Explain the use of a zero-potential location. ● The points on an Equipotential Surface all have the same electric potential. The work done on a test charge in moving it from one such surface to another is inde- pendent of the locations of the initial and final points on these surfaces and of the path that joins the points. The electric field E → is always directed perpendicularly to corresponding Equipotential Surfaces. ● The electric potential difference between two points i and f is V f − V i = − ∫ f i E → · d s → , where the integral is taken over any path connecting the points. If the integration is difficult along any particular path, we can choose a different path along which the integration might be easier. ● If we choose V i = 0, we have, for the potential at a particular point, V = − ∫ f i E → · d s → .

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  We have already encountered one Equipotential Surface. In Section 18.8, we found that the direction of the electric field just outside an electrical conductor is perpendicu- lar to the conductor’s surface, when the conductor is at equilibrium under electrostatic conditions. Thus, the surface of any conductor is an Equipotential Surface under such 19.4 Equipotential Surfaces and Their Relation to the Electric Field 591 conditions. In fact, since the electric field is zero everywhere inside a conductor whose charges are in equilibrium, the entire conductor can be regarded as an equipotential volume. There is a quantitative relation between the electric field and the Equipotential Surfaces. One example that illustrates this relation is the parallel plate capacitor in Figure 19.15. As Section 18.6 discusses, the electric field → E between the metal plates is perpendicular to them and is the same everywhere, ignoring fringe fields at the edges. To be perpendicular to the electric field, the Equipotential Surfaces must be planes that are parallel to the capacitor plates, which themselves are Equipotential Surfaces. The potential difference between the plates is given by Equation 19.4 as ΔV = V B − V A = −W AB /q 0 , where A is a point on the positive plate and B is a point on the negative plate. The work done by the electric force as a positive test charge q 0 moves from A to B is W AB = FΔs, where F refers to the electric force and Δs to the displacement along a line perpendicular to the plates. The force equals the product of the charge and the electric field E (F = q 0 E), so the work becomes W AB = F Δs = q 0 E Δs. Therefore, the potential difference between the capacitor plates can be written in terms of the electric field as ΔV = −W AB /q 0 = −q 0 EΔs/q 0 , or E = − ΔV  ___ Δs  (19.7a) The quantity ΔV/Δs is referred to as the potential gradient and has units of volts per meter.

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  eBook - PDF

  Introduction to Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  The work, accord- ing to Equation 19.4, would cause the potential to change, and, thus, the surface could not be an Equipotential Surface as assumed. The only way out of the dilemma is for the electric field to be perpendicular to the surface, so there is no component of the field parallel to the surface. We have already encountered one Equipotential Surface. In Section 18.8, we found that the direction of the electric field just outside an electrical conductor is perpendicu- lar to the conductor’s surface, when the conductor is at equilibrium under electrostatic conditions. Thus, the surface of any conductor is an Equipotential Surface under such conditions. In fact, since the electric field is zero everywhere inside a conductor whose charges are in equilibrium, the entire conductor can be regarded as an equipotential volume. There is a quantitative relation between the electric field and the Equipotential Surfaces. One example that illustrates this relation is the parallel plate capacitor in Figure 19.15. As Section 18.6 discusses, the electric field E B between the metal plates is perpendicular to them and is the same everywhere, ignoring fringe fields at the edges. To be perpendicular to the electric field, the Equipotential Surfaces must be planes that are parallel to the capacitor plates, which themselves are Equipotential Surfaces. The potential difference between the plates is given by Equation 19.4 as DV 5 V B 2 V A 5 2W AB /q 0 , where A is a point on the positive plate and B is a point on the negative plate. The work done by the electric force as a positive test charge q 0 moves from A to B is W AB 5 FDs, where F refers to the electric force and Ds to the displacement along a line perpendicular to the plates. The force equals the product of the charge and the electric field E (F 5 q 0 E), so the work becomes W AB 5 FDs 5 q 0 EDs.

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  eBook - PDF

  University Physics Volume 2

  • William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
  • 2016(Publication Date)
  • Openstax

    (Publisher)

  A two-dimensional map of the cross-sectional plane that contains both charges is shown in Figure 7.34. The line that is equidistant from the two opposite charges corresponds to zero potential, since at the points on the line, the positive potential from the positive charge cancels the negative potential from the negative charge. Equipotential lines in the cross-sectional plane are closed loops, which are not necessarily circles, since at each point, the net potential is the sum of the potentials from each charge. Figure 7.34 A cross-section of the electric potential map of two opposite charges of equal magnitude. The potential is negative near the negative charge and positive near the positive charge. View this simulation (https://openstaxcollege.org/l/21equipsurelec) to observe and modify the Equipotential Surfaces and electric fields for many standard charge configurations. There’s a lot to explore. One of the most important cases is that of the familiar parallel conducting plates shown in Figure 7.35. Between the plates, the equipotentials are evenly spaced and parallel. The same field could be maintained by placing conducting plates at the equipotential lines at the potentials shown. 322 Chapter 7 | Electric Potential This OpenStax book is available for free at http://cnx.org/content/col12074/1.3 Figure 7.35 The electric field and equipotential lines between two metal plates. Note that the electric field is perpendicular to the equipotentials and hence normal to the plates at their surface as well as in the center of the region between them. Consider the parallel plates in Figure 7.2. These have equipotential lines that are parallel to the plates in the space between and evenly spaced. An example of this (with sample values) is given in Figure 7.35. We could draw a similar set of equipotential isolines for gravity on the hill shown in Figure 7.2. If the hill has any extent at the same slope, the isolines along that extent would be parallel to each other.

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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)

  To illustrate the idea of electrostatic potential, consider a distribution of charges in a plane. Positive charges tend to move toward regions of negative electrostatic potential. Negative charges spontaneously move toward regions of positive electrostatic potential.

  Think of the contours of an electrostatic potential surface in the same way you think of the contours of altitude on a geological map. A geological map plots lines of constant altitude z(x,y) as a function of north–south and east–west coordinates (x,y). Such maps describe the gravitational potential energy field. If you walk from coordinate (x1 , y1 ) to coordinate (x2 , y2 ), your altitude increases from z1 to z2 . In this analogy, the positions of the mountains are fixed and the mass that you carry up the mountain is called the test particle. Multiplying that mass by the altitude difference gives the gravitational energy change: mass × gravitational acceleration constant × height. We do the same calculation here with charges and electrostatic potentials, rather than with masses and altitudes. No work is required to move charges along contours of constant electrostatic potential, just as no gravitational work is done in walking along level paths on a hillside. For a point charge, a surface of constant electrostatic potential is any sphere of radius r centered on the point. It requires no work to move a test charge anywhere on that sphere (see Figure 21.2 ). The electric field E is always perpendicular to an Equipotential Surface.

  Figure 21.2 The electrostatic potential is constant at distance r from a point charge. Lines of equipotential are circles in two dimensions. Broken lines indicate the direction of force and the field E.

  The Equipotential Surfaces Around Two Positive Point Charges

  Figure 21.3 Lines of equal electrostatic potential around two positive point charges.

  The dashed lines in Figure 21.3 show the electric field E in the (x,y) plane due to two charges of the same sign: q at x = -ℓ/2, and q at x = +ℓ/2. The x axis is an axis of symmetry of this electric field and charge constellation. The solid curves indicate the intersection of the plane of the page and the Equipotential Surfaces. If you rotate the solid curves about the x axis, out of the plane of the page, you generate the full three-dimensional Equipotential Surfaces. The Equipotential Surfaces become more nearly perfect spheres as their distance from the charges increases. Therefore, at a distance far away from the point charges, you can view the electrostatic potential as though it were due to a single point charge equal to 2q at x

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  eBook - ePub

  Introduction to Numerical Electrostatics Using MATLAB

  • Lawrence N. Dworsky(Author)
  • 2014(Publication Date)
  • Wiley-IEEE Press

    (Publisher)

  1.28 ) is a scalar equation, which is almost always easier to work with than is a vector equation. Also, once the voltage is known, it is a straightforward job to calculate the field. Consequently, we will concentrate on finding voltages and then (if necessary) finding the field, not the other way around.

  For the single-point charge of equation (1.28 ), we already know that the field lines point radially outward (from the charge), going to infinity. Figure 1.9 . shows surfaces of constant potential, known as Equipotential Surfaces or more commonly equipotentials. These surfaces cross the field lines normally and in this situation are spheres.

  FIGURE 1.9

  Electric field lines and Equipotential Surfaces about a point charge.

  If, instead of a single charge q, we have a collection of (discrete) charges, we must replace equation (1.28 ) by the sum of the contributions of all the charges, and a is replaced by the distances from each of the charges (xi ,yi ,zi ) to the measurement point p = (xp ,yp ,zp ). In other words,

  (1.29)

  and then

  (1.30)

  The gradient [equation (1.26 )] operating on V produces en electric field vector whose direction is the same as that of the maximum change in V. Since the direction of maximum change from an equipotential contour is always normal to the contour, the electric field lines will always be normal to the equipotential contours. Figure 1.10 is essentially a repeat of Figure 1.1

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  Fundamentals of Physics, Extended

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

    (Publisher)

  The work done on a test charge in moving it from one such surface to another is independent of the locations of the initial and final points on these surfaces and of the path that joins the points. The electric field E → is always directed perpendicularly to corresponding Equipotential Surfaces. Finding V from E → The electric potential difference between two points i and f is V f − V i = − ∫ f i E → · d s → , (24-18) where the integral is taken over any path connecting the points. If the integration is difficult along any particular path, we can choose a dif- ferent path along which the integration might be easier. If we choose V i = 0, we have, for the potential at a particular point, V = − ∫ f i E → · d s → . (24-19) In the special case of a uniform field of magnitude E, the potential change between two adjacent (parallel) equipotential lines separated by distance Δx is ΔV = –E Δx. (24-21) Potential Due to a Charged Particle The electric potential due to a single charged particle at a distance r from that particle is V = 1 4πε 0 q r , (24-26) where V has the same sign as q. The potential due to a collection of charged particles is V = ∑ n i =1 V i = 1 4πε 0 ∑ n i =1 q i r i . (24-27) Potential Due to 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 (24-30) for r ⪢ d; the angle θ is defined in Fig. 24-13. Potential Due to a Continuous Charge Distribution For a continuous distribution of charge, Eq. 24-27 becomes V = 1 4πε 0 ∫ dq r , (24-32) in which the integral is taken over the entire distribution. Calculating E → from V The component of E → in any direction is the negative of the rate at which the potential changes with dis- tance in that direction: E s = − ∂V ∂s . (24-40) The x, y, and z components of E → may be found from E x = − ∂V ∂x ; E y = − ∂V ∂y ; E z = − ∂V ∂z .

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  Principles of Physics: Extended, International Adaptation

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2023(Publication Date)
  • Wiley

    (Publisher)

  2. The x, y, and z components of E → may be found from E x = − ∂V ___ ∂x ; E y = − ∂V ___ ∂y ; E z = − ∂V ___ ∂z . When E → is uniform, all this reduces to E = − Δ V ____ Δs , where s is perpendicular to the Equipotential Surfaces. 3. The electric field is zero parallel to an Equipotential Surface. LEARNING OBJECTIVES 704 CHAPTER 24 Electric Potential Calculating the Field from the Potential In Module 24.2, you saw how to find the potential at a point f if you know the electric field along a path from a reference point to point f. In this module, we propose to go the other way—that is, to find the electric field when we know the potential. As Fig. 24.2.2 shows, solving this problem graphically is easy: If we know the potential V at all points near an assembly of charges, we can draw in a family of Equipotential Surfaces. The electric field lines, sketched perpendicular to those surfaces, reveal the variation of E → . What we are seeking here is the math- ematical equivalent of this graphical procedure. Figure 24.6.1 shows cross sections of a family of closely spaced equipo- tential surfaces, the potential difference between each pair of adjacent surfaces being dV. As the figure suggests, the field E → at any point P is perpendicular to the Equipotential Surface through P. Suppose that a positive test charge q 0 moves through a displacement d s → from one Equipotential Surface to the adjacent surface. From Eq. 24.1.6, we see that the work the electric field does on the test charge during the move is –q 0 dV. From Eq. 24.2.2 and Fig. 24.6.1, we see that the work done by the electric field may also be written as the scalar product ( q 0 E → ) ⋅ d s → , or q 0 E(cos θ) ds. Equating these two expressions for the work yields –q 0 dV = q 0 E(cos θ) ds, (24.6.1) or E cos θ = − dV ___ ds . (24.6.2) Since E cos θ is the component of E → in the direction of d s → , Eq.

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