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PHY132 Formulae Midterm 2

University Physics with Modern Physics with Mastering Physics (11th Edition)

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Formulae: Electric Charge Q [C] : Q e = 1.602×10 –19 C = e, Q p = + e, m e = 9.11×10 –31 kg, m p = 1.67×10 –27 kg Current: I dQ/dt Unit: Ampere A C/s I = qn q Av drift ; v drift E q is the charge of the mobile charges, n q their density, A the wire’s cross section, E the local electric field strength Coulomb Force: Force between two point-charges Q 1 and Q 2 separated by distance r 12 Force exerted by Q 1 on Q 2 , r 12 points from 1 to 2 9 2 2 12 2 2 1 2 1 on 2 0 2 0 12 0 1 1 , 8.998 10 Nm /C , 8.85 10 C /(Nm ) 4 4 Q Q r ε πε πε = = × = × F r ± Electric Field (Vector Field!): E F E / q Force-per-unit-test-charge: N/C Field of a point charge Q Field outside a spherically symmetric charge Q ( r>R ) 2 0 force on charge : 1 , 4 q q Q q r πε = = E r F E ± Field of a line charge λ =Q/L Field outside ( r n >R ) a cylindrically symmetric charge λ l l 0 1 ( is unit vector normal to the line) 2 n n r λ πε = r E r Field of a planar charge σ =Q/A ; for infinite plane, or close compared to the size of the plane… l l 0 ( is unit vector normal to the plane) 2 σ ε = A E A Field of a uniformly charged ring, radius R , on a line ring and through the center ( x -axis!) 2 2 3 2 2 2 3 2 0 0 1 4 ( ) 2 ( ) Q R x R x R λ πε ε = = + + x x E Field inside a conductor E =0 when no EMF is present: the mobile charges rearrange them- selves to cancel an external field ( E 0 when an EMF is present!) Energy density u (energy/unit volume) of electric field E : 2 0 2 u E ε = Gauss’ Law for E: 0 closed surface Flux enclosed E Q d ε Φ = E A v Choose the Gaussian surface to match the symmetry of charge distribution and E -field! Magnetic Force: F B = q v×B d F B = I d l×B Force on a charge q , moving with velocity v in magnetic field B Force on a wire of length d l carrying current I , in magnetic field B Force between two parallel wires of length l , car- rying currents I
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