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Notes Lecture for Chapter 19 http://www.physics.drexel.edu/~jelena/Phys102Lectures /Lectures102Post.html 1 Applying Gauss' Law Gauss' Law is valid for the electric field of any system of charges or continuous distribution of charge. Although Gauss' Law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations Particularly spherical, cylindrical, or plane symmetry Remember, the Gaussian surface is a surface you choose, it does not have to coincide with a real surface 2 Conditions for a Gaussian Surface Try to choose a surface that satisfies one or more of these conditions: The value of the electric field can be argued from symmetry to be constant over the surface The dot product can be expressed as a simple r r algebraic produce E dA because E and A are parallel r r The dot product is 0 because E and A are perpendicular The field can be argued to be zero everywhere over the surface 3 Problem 19.37. A cylindrical shell of radius 7.00 cm and length 240 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell. 4 Example 19.9: The Electric Field Due to a Point Charge Choose a sphere as the Gaussian surface E is parallel to dA at each point on the surface r r E = E dA = EdA = q r2 qin o = E dA = E (4 r 2 ) E= q 4 o r 2 = ke 5 Example 19.10: Field Due to a Spherically Symmetric Charge Distribution Select a sphere as the Gaussian surface For r> a qin E = E dA = EdA = o Q Q E= = ke 2 2 4 or r 6 Spherically Symmetric, cont Select a sphere as the Gaussian surface, r < a qin < Q qin = (4/3r3) E = E dA = E= 4 o r 2 qin = EdA = qin o keQ r 3 a 7 Spherically Symmetric Distribution, final Inside the sphere, E varies linearly with r E 0 as r 0 The field outside the sphere is equivalent to that of a point charge located at the center of the sphere 8 Example 19.11: Field at a Distance from a Line of Charge Select a cylindrical charge distribution The cylinder has a radius of r and a length of l E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface 9 Field Due to a Line of Charge, cont The end view confirms the field is perpendicular to the curved surface The field through the ends of the cylinder is 0 since the field is parallel to these surfaces 10 Field Due to a Line of Charge, final Use Gauss' Law to find the field E = E dA = EdA = qin o l E ( 2 r l ) = o = 2ke E= r 2 o r 11 Example 19.12: Field Due to a Plane of Charge r E must be perpendicular to the plane and must have the same magnitude at all points equidistance from the plane Choose a small cylinder whose axis is perpendicular to the plane for the Gaussian surface 12 Field Due to a Plane of Charge, cont r E is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinder The flux through each of end the cylinder is EA and so the total flux is 2EA 13 Field Due to a Plane of Charge, final The total charge in the surface is A Applying Gauss' Law A and E = E = 2EA = 2 o o Note, this does not depend on r Therefore, the field is uniform everywhere 14 Chapter 20 Electric Potential and Capacitance Electrical Potential Energy When a point charge, qo, is placed in an electric field, it experiences a force r r F = qo E The force is conservative To find the work associated with the force, look at ds, an infinitesimal displacement vector that is oriented tangent to a path through space 16 Electric Potential Energy, cont The work done by the electric field is r r r r Fe ds = qoE ds As this work is done by the field, the potential energy of the charge-field system is changed by U For a finite displacement of the charge from A to B, B r r U = UB - U A = -qo E ds A 17 Electric Potential Energy, final r Because qoE is conservative, the line integral does not depend on the path taken by the charge This is the change in potential energy of the system 18 Electric Potential The potential energy per unit charge, U/qo, is the electric potential The potential is independent of the value of qo The potential has a value at every point in an electric field U The electric potential is V qo 19 Electric Potential, cont The potential is a scalar quantity Since energy is a scalar The potential is a property of only the field Not the charge-field system As a charged particle moves in an electric field, it will experience a change in potential B r r U V = = - E ds A qo 20 Electric Potential, final We often take the value of the potential to be zero at some convenient point in the field Sometimes called ground Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field 21 Potential and Potential Energy The potential is characteristic of the field only It is independent of the charge placed in the field The difference in potential is proportional to the difference in potential energy Potential energy is characteristic of the charge-field system Due to an interaction between the field and a charged particle placed in the field 22 Work and Electric Potential The electric potential at an arbitrary point due to source charges equals the work required by an external agent to bring a test charge from infinity to that point divided by the charge on the test particle Assumes a charge moves in an electric field without any change in its kinetic energy The work performed on the charge is W = V = q V 23 Units 1 V = 1 J/C V is a Volt It takes one Joule of work to move a 1 Coulomb charge through a potential difference of 1 Volt In addition, 1 N/C = 1 V/m This indicates we can interpret the electric field as a measure of the rate of change with position of the electric potential 24 Electron-Volts Another unit of energy that is commonly used in atomic and nuclear physics is the electronvolt One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt 1 eV = 1.60 x 10-19 J 25

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