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The problem solving session will be Wednesdays from 12:30 – 2:30 (or until there is nobody left asking questions) in FN 2.212 Torque on the electric dipole r r p qd (electric dipole moment from “-” to “+”) Electric field is uniform in space Net Force is zero Torque is not zero Net (qE )(d sin ) p E (torque is a vector) Stable and unstable equilibrium p E p E Charge #2 Three point charges lie at the vertices of an equilateral triangle as shown. Charges #2 and #3 make up an electric dipole. The net electric torque that Charge #1 exerts on the dipole is +q Charge #1 +q y –q x A. clockwise. B. counterclockwise. C. zero. D. not enough information given to decide Charge #3 Electric field of a dipole E-field on the line connecting two charges 1 1 E keq 2 2 r2 r1 - A + E r2 d 2 p ke E 3 r r1 E-field on the line perpendicular to the dipole’s axis E 2E 2 sin E2 E 2 E2 d r E 2 ke A E1 - q r2 qd E ke 3 r r r p E k e 3 r when r>>d + d General case – combination of the above two 3( p r ) p E r 3 5 r r Dipole’s Potential Energy E-field does work on the dipole – changes its potential energy Work done by the field (remember your mechanics class?) dW d pE sin d U p E Dipole aligns itself to minimize its potential energy in the external E-field. Net force is not necessarily zero in the non-uniform electric field – induced polarization and electrostatic forces on the uncharged bodies Chapter 22 Gauss’s Law Charge and Electric Flux Previously, we answered the question – how do we find E-field at any point in space if we know charge distribution? Now we will answer the opposite question – if we know E-field distribution in space, what can we say about charge distribution? Electric flux Electric flux is associated with the flow of electric field through a surface 1 E~ 2 r S ~ r2 E S const For an enclosed charge, there is a connection between the amount of charge and electric field flux. Calculating Electric Flux Amount of fluid passing through the rectangle of area A dV A dt dV A cos dt dV A dt Flux of a Uniform Electric Field E E A EA cos A A n n - unit vector in the direction of normal to the surface Flux of a Non-Uniform Electric Field E E d A S E – non-uniform and A- not flat Few examples on calculating the electric flux Find electric flux E 2 103[ N / C ] Gauss’s Law E E d A qi 0 Applications of the Gauss’s Law Remember – electric field lines must start and must end on charges! If no charge is enclosed within Gaussian surface – flux is zero! Electric flux is proportional to the algebraic number of lines leaving the surface, outgoing lines have positive sign, incoming - negative Examples of certain field configurations Remember, Gauss’s law is equivalent to Coulomb’s law However, you can employ it for certain symmetries to solve the reverse problem – find charge configuration from known E-field distribution. Field within the conductor – zero (free charges screen the external field) Any excess charge resides on the surface E d A0 S Field of a charged conducting sphere Field of a thin, uniformly charged conducting wire Field outside the wire can only point radially outward, and, therefore, may only depend on the distance from the wire Q E d A 0 E 2 r 0 - linear density of charge Field of the uniformly charged sphere Uniform charge within a sphere of radius r r q Q a 3 ' E r 3 0 Q - total charge Q - volume density of charge V Field of the infinitely large conducting plate s Q A s- uniform surface charge density s E 2 0 Charges on Conductors Field within conductor E=0 Experimental Testing of the Gauss’s Law