Semester II, 2017-18 Department of Physics, IIT Kanpur PHY103A: Lecture # 8 (Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.) Anand Kumar Jha 19-Jan-2018 Summary of Lecture # 7: • Electrostatic Boundary Conditions E E = Electric Field: = above below E − E = 00 above− below � 0 above − below Electric Potential: V V = 0 above − below • Basic Properties of Conductors (1) The electric field = 0 inside a conductor, always. = 0 = = 0 (2) The charge density inside a conductor. This is because . (3) Any net charge resides on the surface. Why? To minimize the energy. 0 ⋅ (4) A conductor is an equipotential. (5) is perpendicular to the surface, just outside the conductor. 2 Summary of Lecture # 7: Prob. 2.36 (Griffiths, 3rd Ed. ): - Surface charge ? = = − 2 - Surface charge ? 4 + - Surface charge ? = − 2 4 1 2 - ( ) ? = 4 out 2 1 0 � - ( ) ? = 4 out 2 10 �+ - ( ) ? =4 out out 2 � - Force on ? 0 40 - Force on ? 0 3 Summary of Lecture # 7: Prob. 2.36 (Griffiths, 3rd Ed. ): - Surface charge ? = Same ª = − 2 Same - Surface charge ? 4 ª + - Surface charge ? = − 2 Changes � 4 1 2 = - ( ) ? 4 Same ª out 2 1 0 � = 4 - ( ) ? Same ª out 2 � 10 + Changes � - ( ) ? =4 out out 2 - Force on ? 0 0 � 4 Same ª Bring in a third - Force on ? 0 Same ª charge 4 Surface Charge and the Force on a Conductor: What is the electrostatic force on the patch? Force per unit area on the patch is: = (? ) = other = + , above other patch above = + 2 other 0 � = + , below other patch below 1 = = + 2 2 other − � other above below 0 But, inside a metal, = , so = 0 below = = + = 2 2 other � above other � � 5 0 0 0 Surface Charge and the Force on a Conductor: What is the electrostatic force on the patch? Force per unit area on the patch is: = (? ) = other = = 2 other � above � Force per unit0 area on the patch is: 0 = = 2 2 other � Force per unit area is pressure.0 So, the electrostatic pressure is: = = 2 2 20 2 6 0 Capacitor: Two conductors with charge and . − What is the potential difference between them? ( ) = = + . ( ) + − − − � − Capacitance is defined as: • Capacitance is≡ the ability of a system to store electric charge. • It is purely a geometric quantity. • C is measured in farads (F), Coulomb/Volt. • Practical units are microfarad (10 ) or picofarad (10 ). −6 −12 7 Work needed to charge a Capacitor: Two conductors with charge and . − How much work needs to be done to increase the charge by d Recall The work required to create a system of a point charge : = QV = = The work necessary to go from = 0 to = is 1 1 = = = 2 2 2 2 � 0 8 Capacitor: Ex. 2.10 (Griffiths, 3rd Ed. ): Find the capacitance of a parallel plate capacitor. Area = A, Separation = The electric field between the plates E = + = = 2 2 0 0 0 0 The potential difference is therefore, = . = = − � 0 Capacitance is: = = 0 9 Capacitor: Ex. 2.11 (Griffiths, 3rd Ed. ): Find the capacitance of two concentric spherical metal shells, with radii and . Suppose there is charge on the inner shell and on the outer shell. − The electric field between the two shells is = 2 � The potential4 difference0 is therefore, = = . = 2 − − � − � 0 1 41 1 = = 2 − 0 � 0 − 4 4 Capacitance is: = = ( ) 10 40 − Superposition principle for electrostatic energy: We have seen several electrostatic systems, including conductors. We know how to calculate electrostatic energy for different system. We know that electric field ( ) and electric potential ( ) follow the principle of superposition. = + + = V + V + Does electrostatic energy also ⋯follow the principle1 2of superposition?⋯ No Why? Because is quadratic in ? = = ( + ) = ( + + 2 ) 2 2 20 0 2 0 2 2 2 � � � 1 2 ⋅ = + + 2 2 0 2 0 2 � 1 � 2 0 � ⋅ = + + 1 2 0 � ⋅ 11 Electric Field: Can We See It? https://www.youtube.com/watch?v=7vnmL853784 12 Special Techniques (Electrostatics): How to find ? = V V − V = − � ⋅ 13 ∞ Special Techniques (Electrostatics): • Laplace’s Equation • The Method of Images • Multipole Expansion 14 Laplace’s Equation Q: How to find electric field ? 1 Ans: ( ) = r (Coulomb’s Law) r Very difficult� 2to calculatê the integral except for very simple situation 40 Alternative: First calculate the electric potential 1 V( ) = r This integral� is relatively easier but in general still difficult to handle 40 Alternative: Express the above equation in the different form. V = (Poisson’s Equation) 2 − When = 0 V0 = 0 (Laplace’s Equation) 2 If = 0 everywhere, = 0 everywhere If is localized, what is away from the charge distribution? 15 Laplace’s Equation in One Dimension V = 0 (Laplace’s Equation) 2 In Cartesian coordinates, + + = 0 2 2 2 2 2 2 If V(, , ) depends on only one variable, , We have = 0 (One-dimensional Laplace’s Equation, 2 ordinary differential equation) 2 General Solution: = + How to calculate the constants and ? Using boundary conditions What decides the boundary condition? The charge distribution 16 .