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potential%20notes%20Sp%2010

# potential%20notes%20Sp%2010 - Electric forces and...

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Unformatted text preview: Electric forces and gravitational forces have many similarities, one of the most important is that both are conservative. As a result there must be an electric potential energy UE associated with FE, just as there is a gravitational potential energy Ug = mgy due to PE. Consider a positive test charge q0 placed in a vertical electric ﬁeld. It experiences a downward electric force F = qu, if the charge is moved upward through a distance d, the electric force and the displacement are in opposite directions, therefore the work done by the electric force is negative. W = -qud Using the deﬁnition of potential energy , i.e. AU = -W we get AU = -W = qud N_ot_e_: that UE increases just as Ug when it is raised against the ﬁeld, on the other hand if qD is negative the FE acting on it will be upward, in this case the FE does positive work as the charge is raised through the distance d and the change in potential energy is negative. Thus the change in potential energy depends on the sign of the charge as well as on its magnitude. The electric force depends on q, the same way as does the change in electric potential energy —‘. . "" .. AU —W 1.e.E=——1t1susefultouseAV=—e= go go go where the SI unit of V is the volt 1 volt = lJoUle/Coulomb As with Ug we deal only with the changes in potential AV, i.e. V can be set to zero at any desired location, just as height can be set to zero in gravitational problems. In addition, both V and U are scalar quantities, so combining them is a matter of simple algebra. There is a connection between E and V that is both straight forward and useful. To obtain this relation apply the deﬁnition AV = -W/qo to the case of a test charge that moves through a distance As in the direction of the ﬁeld. The work done by the electric ﬁeld is the force times the distance W = qud , therefore the change in electric potential is AV = —W/qo = ~ (q.,Ed)/qo = — EAs Solving for the ﬁeld E = -AV/As this shows that the ﬁeld which can be expressed in terms of Force/charge i.e. N/C also has units of Volts/meter i.e. l N/C = l V/m Therefore if the potential (V) is known then the electric ﬁeld (E) can be found, or if E is known, (such as ﬁom Gauss’s Law) then V can be found To summarize: the electric ﬁeld depends on the rate of change of the electric potential with position (in terms of gravity you can think of the potential V as the height of a hill, and the electric ﬁeld E as the slope of the hill) in addition it follows that V decreases as one moves in the direction of E (i.e. V = - EAs is ' negative when both E and As are in the same direction) Example: If you have a constant potential in a region of space then what is the electric ﬁeld in this region? (a) positive, {23) negative, (c) zero (the answer is 0 zero) Reasoning: since E is the rate of change of V with position and the rate of change of a constant potential is zero, so must be the ﬁeld. Energy Conservation: when a ball is dropped in a gravitational ﬁeld its Ug decreases as it falls, at same time its KE increases. If non-conservative forces are ignored then the energy is conserved AUdecrease = AKEincrease Because F E is also conservative the same considerations apply to a charged object in an electric ﬁeld i.e. KEA'l‘ PEA = KEB + PEB Recall that if forces are due to gravity PE = Ug = mgy springs PE = Usp = V2 lot2 electric PE = qV A h‘.’ \V example: if an object of mass m = 1.75 x 10'5 kg and charge q = 5.2 x 10‘5C is released from point A at rest, as it moves to point B its electric potential decreases by 60Volts (i.e. VA-VB = 60V) the speed of the object at point B is? KEA + PEA = K13B + FEB 1/2 mvb2 = 1/2 mVa2 + q(VA'VB) vB = \l 2q(VA-VB)/m = 13.9 m/s therefore a decrease in V appears as an increase in KB, if a negative charge is used the situation is the opposite. A negative charge will move to a region of higher electric potential with an increase in speed, i.e. if q = -5.2 x 10'5C and AV = -60V, the ﬁnal speed is the same in general, positive charges accelerate in the direction of decreasing V negative charges accelerate in the direction of increasing V in both cases however, the charge moves to a region of lower electric potential energy example: an electron is released and accelerates through V, a proton is released and accelerates through —V, the ﬁnal speed of the electron is (a) faster, (b) slower, (c) the same as the proton? The answer is (a) faster, since masses are different, but there is still the same magnitude change in potential V for a point charge: Consider a point charge +q that is ﬁxed at the origin of a coordinate system Suppose a test charge +qo is held at rest at point A, here it experiences a force F = quo/rAZ. If the test charge is now released it will accelerate away from the origin when it reaches point B its KE will have increased by the same amount as the decrease in electric potential energy _ quo quo i.e. UA is greater than U3 and UA —UB — r A r B 1 k and VA- VB = — (In-U3)= 5"— -—q go rA ’19 if we choose r = no as a reference point then U = O at this point qu therefore V = — for a point charge and U = — for a pair of point charges separated by r r r Superposition of V and U If more than one point charge exists in a system, then use superposition principle to ﬁnd the total potential of the system. V = ZVi where Vi = E The total work is the sum of the work done due to the individual r k forces. Wt = 2W; = Z qq” = qAV r Example: if q = 4.11 x 10’9C is at O m and -2q is at 1m then, (a) what is V at the midpoint +53 - ‘ of the between the charges, and (b) where does V = O _ k —2 ‘ - ' ' 7a solution (a) V = E + —(—-q—)- ifx = 0.5 m then x (1 — x) V = kq/O.5 + k(-2q)/(1—O.5) = -73.9 Nm/C = -73.9 V (b) V = —kq/x + k(—2q)/(1-x) = O l/x =2/(l—x) x = 0.33m and V = 0 at x = -1m r I I / / 9“ i‘ "elm 0 Im If the system is a continuous object, then break the object into parts, with each element having a charge dq, ﬁnd the element of potential due to dq , and then integrate V: “W = j 73 Equipotential surfaces and electric ﬁeld ( I. A contour map is a useful tool, it is a series of closed curves, the contours each denoting a different altitude, . - When the contours are closely spaced the altitude changes rapidly indicating steep terrain, widely spaced ' Kg... .. contours indicate a fairly ﬂat surface. A similar device can help visualize V * "‘ Any surface for which all points are at the same potential is called an equipotential surface decreasing in magnitude with Increasing distance from the source charge from the source charge Note: The choice of speciﬁc potentials is arbitrary, an equipotential surface can be drawn through any point in the ﬁeld. They are perpendicular to the electric ﬁeld, if it were not perpendicular then motion of f/ charges along the surface would correspond to a change in PE due to work done by the ﬁeld on the charge. If excess charge is on a conductor it resides on the surface, if the charge is at rest, whatever E is present must be perpendicular to the surface, since any ﬁeld parallel to the surface would cause the charge to move. Because the ﬁeld is perpendicular to the surface all points along the surface must be at the same potential, therefore, when all charges have come to rest, the surface of a conductor is always an equipotential surface, and inside the conductor V = constant (that of the surface) Problem solving hints for potentials 1. Recall potentials are scalar quantities, therefore add algebraically for a discrete system of charges (Keep track of signs) The potential is positive for positive charges and negative for negative charges. 2. The choice of zero potential is arbitrary, since only changes in potential are signiﬁcant (just like mechanics) 3. To ﬁnd the electric potential at some point P due to a continuous chﬁarj distribution break into arts and then inte ate ‘ I, ' - i p gr wise/e pol/I . ' I” 4. If E is known, then V can be found h . e— 00 5. If V is known then B can be found since E is equal to the negative derivative of potential with respect to some coordinate «,4 . .1 v ~ 3? V j" _ .gv ‘ 11‘: .«cpv i Ex», v E- GWEN)» I ._..__-_ Pr 37‘ / J d V Exam???” pr STIMié )l't Iﬂnwcfplﬂﬂ 1'5 ﬂ'7aL9ﬁO'i‘L/Uiﬂe pie/45,60)”; W7 (“7(1) ﬁfﬂmb a) Ell/760217er Ope'ujia Azg‘éﬂC/m » W V7667,” 1,5 M fofewfi‘aJ lHewev’Cd’ jo'ﬂi-QEQ—Ol— M («gin/Q pk F140;) 5%? Ci (ibfﬂmcz d) l‘ 2‘21’1‘7 éQ/UL / ¥novh®ausslkw ﬁ)‘; A n {format/ire) / I: {775} II] , 7— a '9 Q n BV301 £10K) :- “j . A (9/ c ->‘\ ’ﬂr-l fﬁ‘giﬂéuf £7765 IV) [73/ (bun y‘lqetgmg rm; ,1 ,Qpc [in dljrgeli'om r [3:117 / ~ ' 1 ” ,. '\ , r ', l . — giéO/OOOVUHb (ntgﬁtlim Sit/1Q wire LS "7L Omani ﬂmwg‘ . LU ill r1 Ci )bUaaiJ [ii/7 (i {FE raj/inf) (:72 /U%.€/L [’30 I‘m/titre? ) " ' ...
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