Electric Potential: Calculating Electric Field and Potential Energy, Study notes of Physics

Download Electric Potential: Calculating Electric Field and Potential Energy and more Study notes Physics in PDF only on Docsity! 1 Lecture 21 Electric Potential • Review • Potential due to a Continuous Charge Distribution • Calculating the Field from the Potential • Electric Potential Energy of a System of Point Charges • Potential of a Charged Isolated Conductor 2 Review • Potential from the Electric Field • Potential due to a Point Charge • Potential due to a group of charges • Potential due to an Electric Dipole r qV 04 1 πε = ∑∑ == == n i i i n i i r qVV 101 4 1 πε )0( =⋅−= ∫ i f i VsdEV r r 2 0 2 0 cos 4 1cos 4 r p r dqV θ πε θ πε == 5 Charged Disk • Potential on the central axis (Fig 24-13) ( ) ( ) ( )zRzRz dRRdVV Rz dRR r dqdV dRRdq R −+= + == + == = ∫ ∫ 22 00 2/122 0 2/122 00 2' '' 2 ' )')('2( 4 1 4 1 )')('2( ε σ ε σ πσ πεπε πσ 6 P 24-22 A plastic rod with a charge of -25.6 pC has been bent into a circular arc. Of radius 3.71 cm and a central angle of 120o. With V=0 at infinity what is the electric potential at point P ? The potential is (in SI units) We note that the result is exactly what one would expect for a point-charge –Q at a distance R. This “coincidence” is due, in part, to the fact that V is a scalar quantity 9 12 2 rod rod 0 0 0 1 1 (8.99 10 )(25.6 10 ) 6.20 V. 4 4 4 3.71 10P dq QV dq R R Rε ε ε − − − × × = = = = − = − ×∫ ∫p p p 7 Calculating the Field from the Potential • Electric potential is a scalar: easier to calculate • How to obtain electric field? (Fig 24-14) Write the work done two ways s VE sdEqdVq r r rr ∂ ∂ −= ⋅=− 00 10 • Consider the following electric potential: V(x, y, z) = 3x 2 + 2xy − z2 yx x VEx 26 −−=∂ ∂ −= x y VEy 2−=∂ ∂ −= z z VEz 2=∂ ∂ −= • What electric field does this describe? ... expressing this as a vector: • Something for you to try: Can you use the dipole potential to obtain the dipole field? Try it in spherical coordinates ... you should get (see Appendix): ( ) θθθ πε ˆsinˆcos2 4 2 3 0 += r r aqE r E from V: an Example ẑzŷxˆ)yx(E 22x26 +−−−= r 11 allows us to calculate the potential function V everywhere (keep in mind, we often define VA = 0 at some convenient place) 0q WVV ABAB ≡− ⇒ ∫ •−=− B A AB ldEVV rr If we know the electric field E everywhere, VE ∇−= rr allows us to calculate the electric field E everywhere If we know the potential function V everywhere, • Units for Potential! 1 Joule/Coul = 1 VOLT The Bottom Line 12 Sample Problem 24-5 The electric potential at any point on the central axis of a uniformly charged disk is given by Eq. 24-37, Starting with this expression, derive an expression for the electric field at any point on the axis of the disk ( )zRzV −+= 22 02ε σ Circular symmetry Ex = Ey =0 z VEz ∂ ∂ −= ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + −=−+−= ∂ ∂ −= 22 0 22 0 22 0 1 2 12 2 1 22 Rz z Rz zzRz dz d z VEz ε σ ε σ ε σ Obs. Look back to Eq. 22-26 15 Electric Potential Energy (cont) • Take a two point charge system (Fig 24-15) • Use potential to calculate work • One can use similar arguments for more than two charges r qqVqWU r qV 21 0 2 1 0 4 1 4 1 πε πε === = 16 Sample Problem 24-6 Fig. 24-16 shows three point charges held in fixed positions by forces that are not shown. What is the electric potential energy U of this system of charges. Assume d = 12 cm and that q1 = +q, q2 = -4q, and q3 = +2q, in which q = 150 nC. r qqU 21 04 1 πε = d qqU 21 0 12 4 1 πε = d qq d qqUUWW 32 0 31 0 23132313 4 1 4 1 πεπε +=+=+ J m CCmN d q d qq d qq d qqUUUU o o 2 29 29 2 231312 107.1 12.0 )10150)(10()/1099.8(10 4 1 )2)(4()2)(()4)(( 4 1 − − ×−= × ⋅×=−= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−+ ++ + −+ =++= πε πε 17 Conductors • Claim The surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential). • Why?? If surface were not equipotential, there would be an electric field component parallel to the surface and the charges would move!! + + + + + + + + + + + + + + 20 How is the charge distributed on the cavity wall? (a) Uniformly (b) More charge closer to -q (c) Less charge closer to -q 1B Problem The induced charge will distribute itself nonuniformly to exactly cancel everywhere in the conductor. The surface charge density will be higher near the -q charge. E r -q 21 How is the charge distributed on the outside of the sphere? (a) Uniformly (b) More charge near the cavity (c) Less charge near the cavity 1C Problem As in the previous example, the charge will be uniformly distributed (because the outer surface is symmetric). Outside the conductor the E field always points directly to the center of the sphere, regardless of the cavity or charge. Note: this is why your radio, cell phone, etc. won’t work inside a metal building! -q 22 Charge on Conductor Demo • How is the charge distributed on a non- spherical conductor?? Claim largest charge density at smallest radius of curvature. • 2 spheres, connected by a wire, “far” apart • Both at same potential But: )/( )/( 2 2 LL SS L S rQ rQ ≈ σ σ L S L S L L S S r r Q Q r Q r Q ≈⇒≈ 00 44 πεπε r S r L ⇒ S L L S r r ≈ σ σ Smaller sphere has the larger surface charge density ! Isolated conductor in an external Electric Field — Lal —-— —_— —— —— = SS —— —— — _— a [i —— —_ — — 25 26 Induced charge distribution on conductor via “method of images” • Consider a source charge brought close to a conductor: + - - - - - + + + + + • Charge distribution “induced” on conductor by source charge: • Induced charge distribution is “real” and sources E- field that is zero inside conductor! – resulting E-field is sum of field from source charge and induced charge distribution – E-field is locally perpendicular to surface + + + + + + + + + + - - - - - • With enough symmetry, can solve for σ on conductor – how? Gauss’ Law o surface surfacesurfacenormal r rErE ε σ )( )()( r rrr == 27 Induced charge distribution on conductor via “method of images” • Consider a source charge brought close to a planar conductor: • Charge distribution “induced” on conductor by source charge – conductor is equipotential – E-field is normal to surface – this is just like a dipole o surface surfacesurfacenormal r rErE ε σ )( )()( r rrr == • Method of Images for a charge (distribution) near a flat conducting plane: – reflect the point charge through the surface and put a charge of opposite sign there – do this for all source charges – E-field at plane of symmetry - the conductor surface determines σ. + - - - - - -