Electric Potential and Energy

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Sheena Beasley
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3 Electic Potential and Enegy Te polem: A solid spee of te adius R is omogeneously caged wit te cage Q and put inside an infinite ollow cylinde. Te cylinde inne and oute adii ae a and, R < a <. Te cylinde is omogeneosly caged wit te cage density ρ. Te cente of te spee is on te cylinde axis. Find te potential in te wole space. Hint: Gauss law and supeposition. Te solution: Fist we find te field of te cylinde: de d = 4πkρ in te cylindical coodinates, ϕ, z). We ave E = 4πk, < a ρ d = πkρ a )/, a < < πkρ a )/, > Te potential is calculated as in te pevious polem: πkρ a ) ln/), φ c = E d = πkρ[ )/ a ln/)], πkρ[ a )/ a ln/a)], > a < < < a ) ) 3) Attention: Hee we cannot put φ = fo since te cylinde is infinite and cages ae pesent at aitaily lage distances fom te coodinate oigin: indeed φ wen. We, teefoe, aitaily cose φ = at =. Fo te spee we ave in speical coodinates R, θ, φ), R = z + ) { kq/r, R > R E R = kqr/r 3, R < R 4) Te potential of te spee { kq/r, φ s = kq/r + kqr R )/R 3, Te total potential is φ = φ c + φ s. R > R R < R 5)

4 Te polem: Electic potential Sumitted y: I.D A ing of adius R is unifomly caged wit cage Q. Te ing is placed in te x-y plane, suc tat its cente is at te oigin.. Wat is te potential along te z-axis?. Estimate te potential a distance ɛ fom te z - axis say, in te y-diection), to leading ode in ɛ. Guidance: fist, wite down an exact integal expession fo te potential. Next, expand te agument in te integal as a powe seies in ɛ. Ten, evaluate te integal to leading ode in ɛ. Te solution:. Te cage density on te ing is λ = Q πr, teefoe, we get dq = λrdθ = Qdθ π ) Te distance etween point on z-axis and te cage dq is = R + z. Te potential is: kdq φ = = kq π R + z π dθ = kq R + z ). et us take any dq on te ing wit cage density λ as aove. Ten is a vecto to te cage dq and is defined y = R cos θ, R sin θ, ). is a vecto to te point we seac te potential fo and is defined y =, ɛ, z). Teefoe te distance etween te two vectos is = R cos θ + ɛ ɛr sin θ + R sin θ + z = R + z + ɛ ɛr sin θ 3) Because ɛ is small, ɛ can e ignoed. Teefoe, te potential is: dq π φ = k = k λrdθ R + z ɛr sin θ = By expanding Teylo s seies we get: kλr π R + z dθ 4) R sin θ ɛ R +z = ) Fom 4) and 5) we conclude: φ = kλr π R + z [ + ɛ R sin θ And finally we can wite te potential φ as: R + z ɛ R sin θ R + z ) φ = πkλr R + z + 3 kλr 3 π kq 8 ɛ = R + z ) 5 R + z + 3 kqr ɛ 6 R + z ) 5 ] dθ 6) 7)

5 Electic Potential Te polem: A tin wie of a lengt is positioned along te x-axis / < x < /. Te wie is caged non-unifomly wit te cage density λ = λ x.. Wat is te total cage on te wie?. Wat is te electic potential on te x-axis fo x > /? Te solution: Since y definition dq = λdx, te total cage is / / x Q = dq = λdx = λ dx = ) / / Te electic potential at = x,, ) due to a cage element dq at = x,, ) is dq dϕx) = k = kλ x dx x x ) Te te electic potential is ϕx) = dϕx ) = kλ Finally = kλ = kλ ϕx) = kλ / / / / / / x dx x x = kλ + x ) x x + x ln x / ) x + / x dx x x 3) / / dx = kλ ) x x + x x dx 4) x + x ln / x ) 5) / x 6)

6 Electic Potential Sumitted y: I.D Te polem: A coil of a adius R and te distance etween loops is caged unifomly wit linea cage density λ. Te top end of te coil is located at +Z and its ottom end at Z. Wat is te electic potential at te oigin, if te potential at te infinity te is zeo? Te coil s position vecto is given as: = R cosθ)î + R sinθ)ĵ + θ π ˆk Te solution: et e te position of a cage element dq on te coil and is te oigin = R cos θ, R sin θ, θ π ) ) =,, ) ) = Te potential is kdq φ) = R + θ 4π 3) We aive at d y diffeentiating d = R sin θ, R cos θ, θ ) dθ 5) π So tat d = R + π ) dθ 6) dq = λdl = λ d ) = λdθ R + 7) π We aive at te integal s oundaies y extacting θ fom z and applying te appopiate oundaies of Z z = θ π θ = πz ; θ = πz Now applying all te aove to te initial equation fo te potential φ) = θ θ kλ R + π ) dθ R + θ 4π = kλ R + π ) π πz πz 4) 8) dθ 9) 4π R + θ

7 and noting tat dx = lnx + a a +x + x ) = πkλ R + 4π π ) lnθ + R + θ ) fo simplification we add + ln Finally: = πkλ R + π ) ln q πz + 4π R + πz q ) πz + 4π R + πz πz + πz + 4π R πz πz ) ) )) = in ode to use a + ) a ) = a ) + πz 4π R ) + ln + πz ) πz + πz + 4π R + πz ) )) ) 4π R + πz ) = πkλ R + π ) ln πz 4π + R + πz ) ) ) ) 4π R = πkλ R + π ) ln Z R + Z R + ) = 4πkλ R + π ) ln Z R + Z + ) 3) R φ,, ) = 4πkλ R + ) ) Z ln π R + Z R + 4)

8 Electic Potential Te polem: An infinite -D cystal stuctue is uilt of point cages along te x-axis wit te same distance etween te sequential cages. Te sign of te cages canges fom eac cage to te following one. Te asolute value of eac cage is q. Wat is te potential enegy U of one cage? Hint: ln + x) = x x + 3 x3 4 x Te solution: et us look at some say) positive cage at te point P and calculate te potential at tis point due to te ote cages kq ϕ = n= q ) n = k = kq ) n n n n= n= = kq )... = kq ) +... Te potential enegy is = kq ) ) ln 3) U = qϕ = ln kq 4)

9 Electic Potential Sumitted y: I.D Te polem: Tee ae 3 conducting spees wit te same cente. Tei adii and cages ae R, R, R 3 and Q, Q, Q 3, espectively. In addition R = R, R = R, R 3 = 3R and Q = Q, Q = Q, Q 3 = 3Q.. Find te electic potential in te wole space. Wat would e te cages on te spees if te spees and 3 ae connected wit a tin conducting wie? Te solution: ϕ) = Q R + Q R + Q 3 R 3 Q + Q R + Q 3 R 3 Q + Q + Q 3 = Q R Q R + 3Q 3R = Q R, = Q Q R + 3Q 3R = Q, < R R < < R R < < 3R R 3 = Q Q + 3Q 3R = Q + Q R, Q + Q + Q 3 = Q Q + 3Q = Q, > 3R ) If te spees and 3 ae connected ten te potentials on tem ae equal. Te new cages Q 3, Q on tem ae suc tat Q 3 + Q = Q 3 + Q = 4Q ) ϕ = ϕ 3 3) Te potentials ae ϕ = Q R + Q R + Q 3 R 3 = Q R Q R + Q 3 3R ϕ 3 = Q + Q + Q 3 R 3 = Q + Q 3 Q 3R 4) 5) Solving te equations fo Q 3, Q we get Q = Q Q 3 = 7 Q 6) 7) and te cage on te spee does not cange.

10 Enegy of a disc and a od Sumitted y: I.D Te polem: A disc of a adius R is caged unifomly wit cage density σ. A od of a lengt is caged unifomly wit cage density λ. Te od is pependicula to te disc wic is in te x y plane) and positioned on te axis of symmety of te disc. Te cente of te od is at z >.. Calculate, fom te diect integation of te field, te foce etween te ojects.. Wat is te flux fom te od tat is passing toug te disc? 3. Find te mutual enegy of te two ojects and deive fom it te expession fo te foce. Te solution:. Te foce etween te ojects et, e te positions of cage elements on te disc and te od, espectively. = cos θ, sin θ, ) ) =,, z) ) = = cos θ, sin θ, z) 3) Because of te symmety of te polem te foce is in te z diection only. Te electic field due to te cage element dq on te disc is de z = k dq z 3 dq = σda = σddθ Te electic field of te disc is E z = R π = πkσz R Te foce acting on te od is F = z+ z kσddθz cos θ) + sin θ) + z ) = kσz 3 ) d = πkσ + z ) 3 E z ẑλdz = πkσλ + R + R ) z R + z d + z ) 3 Te foce acting on te disc is te same ut in te opposite diection. π 4) 5) dθ 6) 7) z ) R + z + ) ẑ 8). Wat is te flux fom te od tat is passing toug te disc? Te foce acting on te disc is F = Eσda = σ E z da Te equested flux is Φ = E z da 9) )

11 Ten Φ = F σ = πkλ + R + z ) R + z + ) ) 3. Te mutual enegy of te two ojects Te potential due to te disc on te z-axis is z > ) φz) = πkσ R + z z) ) Te mutual enegy of te disc and te od is z+ U = λ φz )dz = πkσλ z But witing in a diffeent way Ten U = z+ z z+ z dz R + z z ) 3) dz fz ) 4) F z = du dz = fz = z ) fz = z + ) 5) = πkσλ + R + z ) R + z + ) 6)

Physics Courseware Electromagnetism

Physics Courseware Electromagnetism Pysics Cousewae lectomagnetism lectic field Poblem.- a) Find te electic field at point P poduced by te wie sown in te figue. Conside tat te wie as a unifom linea cage distibution of λ.5µ C / m b) Find