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PHY212 Lecture 7
1)2006 exams/solutions is at http://physics.syr.edu/courses/PHY212.07Fall/Quizzes Exams.htm 2)You may bring a formula sheet(written on one page only) to the exam (Exam1 Thursday Sept 20).
- Review mechanics Net Force (F) on point object of mass m/acceleration(a) of object ;
Newton’s law(s) : F = ma = md dtv , v = d dtx −→
work/KE relation : W =
∫ (^) f i F^ ·^ dx^ =^ Kf^ −^ Ki^ =^
1 2 mv
2 f −^ 1 2 mv
2 i
In general, F = − ∂U∂x + (Non-gradient Force), U = potential enery(PE).
- Some reasons for introducing work, potential energy and potential:
- They are scalar quantities and so easier to deal with.
- These concepts may be intuitively closer to everyday communication.
- Gravitational force/field: Force/PE : F = − ∂U∂x
Field/Potential : F/m = −∂(U/m ∂x ) −→ g = − ∂V∂x
work/KE relation becomes Ki + Ui = Kf + Uf = Total energy.
This can also be written as W = ∆K = −∆U
- Electric force/field (Analogous to the gravitational force/field ) Force/PE : Fe = −∂U ∂xe (defining relation for electrical PE)
Field/Potential : Fe/q = −∂(U ∂ex/q ) −→ E = −∂V ∂xe
Ve = Ue/q ⇒ units = J/C=volts(V).
- Acceleration of charges by a potential difference between two parallel plates: Use W = ∆K = −∆U.
- Calculating the electric potential(V ) from the electric field (E). First find E (e.g. by Gauss’ law) then E = −∂V ∂xe −→
V (x) − V (xref ) = −
∫ (^) x
xref
E(x′) · dx′
For a point charge Q located at xQ, E(x) = kQr 2 ˆr, r = |x − xQ|, k = (^4) πε^10.
V (x) − V (xref ) = −
∫ (^) x xref
kQ r′^2 dr
′ (^) = kQ r −^
kQ rref ,^ r^ =^ |x^ −^ xQ|. If we take the reference point xref to be at ∞ where limr→∞ kQr = 0, then V (x) − V (∞) = kQr. Therefore, relative to a point at ∞, the electric potential (V ) at a point x due to a point charge Q located at xQ is V (x) = kQr , r = |x − xQ|.
- Calculating the electric potential(V ) of a charge distribution directly from the electric potential of a point charge. - V at a position x due to a point charge Q located at a different position xQ is defined to be V (x) = kQr , r = |x − x 1 |. The electric potential energy of another point charge q (call it a test charge) now placed at position xq is U = qV (xq) = kqQr , r = |xq − xQ|. - V at position x due to a collection of point charges Q 1 , Q 2 , ... located at respective positions x 1 , x 2 , ... is given by
V (x) =
kQ 1 r 1
kQ 2 r 2
i
kQi ri
, ri = |x − xi|.
Again, the electric PE of another point charge q now placed at position xq is U = qV (xq) = kq
i
Qi ri ,^ ri^ =^ |xq^ −^ xi|.
- V at the point x due to a continuous charge distribution, which is con- tained in a certain region R in space, is given by
V (x) =
R
kdQ r
, r = |x − xdQ|.
where the charge element dQ is treated like a point charge. The electric PE of a point charge q place at position xq is
U = qV (xq) = kq
R
dQ r
, r = |xq − xdQ|.
- Finally, just to be complete, the electric PE of a collection of point charges Q 1 , Q 2 , ... located at respective positions x 1 , x 2 , ... is given by
U =
i 6 =j
kQiQj rij
, rij = |xi − xj |.
Similarly, the electrical PE of a continuous charge distribution occupying some region R of space is
