Study Guides (248,531)
Canada (121,614)
Physics (82)
PCS 211 (23)
Final

PCS 211 Final: Chapter 25
Premium

9 Pages
92 Views
Unlock Document

Department
Physics
Course
PCS 211
Professor
Alexandre Douplik
Semester
Winter

Description
• • pointhan▯e in PoForc=:=FEnqrgy by movinGravitatiFolrce25.1Photntial5iffectcecnodeltitlic Potentia B The The integral is called a on the path taken from . This is an essential property of conservative forces. : B A g ▯ ▯ U W = = mg F S ▯ ▯ Conservative Forces path integraS S = = q mg 0▯ ▯ AEB S • or adS ▯ ▯ U W B A line integra= = q ▯ F E l from0point E ▯ S 0 S = . q S A E and A S to elecrAcfied:ivalent un*Unit of popential: voltThe potentiaDefineerence: 1 V is the work required per un V s aitalar.otential energy per unit charge. potcatclla.te E from = Electric potential 1 C1 J Useful to V C N p = = m V ▯ ▯E P • ds : V= p it charge to bring a positive test . ▯ V ▯ 0 at = U/q 0q U ▯ = 0. ▯ A B E ds *Notice the difference between potential and potential energy.tial Di1etetcal.EieeUrnofV(flt(eltri)gaelrsy common energy unit: the energy Electricfiesieikimalunet.ergy increases, the potential energy decreases ▯ V = : Release a positive charge in a uniform field. BV 0 ▯ AV ▯ U = q = 10 ▯ -19 A B ways point to lower potential direction. C V =1.60 ▯ • V U ds = = q ▯ or loses by moving through a ▯ A B ▯ V E 10 = cos0 J19 q El ds = ▯ El Equipotential Surface lines.erpendichlerythheee.eIimiufsedtial is the is a surface on • • • • d=0.3cm kineIifcaneroynw|E|? WhiDhirsdtihaofhtheeerpotintfiiaell?d?its E Example: The electric field between two parallel plates = V B l V A = | 0.3 + + + + + + ▯ 12 10 V 2 m 12V = - ▯ - - - - - - 10 3 V m (This For multiplNote,Tntch,argeTherefFooePGenerally,e:tw(3)nFiointorant)D.defrerDera3uileceiitflledtines. Potential Energy Due to Point Charges ▯ V depends only on and point ▯ ˆ ▯ E ▯ V • V = V ▯ r = ds = k = B 0 atAand ▯ = ▯ re V . A▯Br ds A▯ B ˆq ▯ A r e cos e V B = ,not on the path. q A ▯ dr ▯ • = ds . Then = dr ds ▯ Br k AE B V q • r = B[1] ds k V ▯ ▯ = A 1 A A i r q k i i r q (b) Find the work re(a) Find V2.Pm3 a 3 -6xample: The potential Due to Two Point Charges W pV pV = = = 3q 8.99 ek ) ) pV ▯ ▯ ▯ xm4 C C = ▯ 1r 1 (3 10 + ▯ 9 2r 2 10 C Nm ▯ ▯ ▯ ▯ 2 2 C6 ▯ ▯▯2▯ )( ▯ ▯ m 10 6.29 ▯ C6 ▯ ▯ 10 6 V3 C charge from ) m 10 = 6 18.9 C ▯ ▯▯ ▯ ▯ 10 6.29 ▯ ▯ 3 to P. ▯ J 10 3 V Together: (b) Second, briThree Charges:n charge q fromEnerTgwoPotential Energy of a Charge Particle Systemd charge q must be done.f q to the system to bring the two charges near each other. ▯ q q to the point P without acceleration: 2< 0, the>0,
More Less

Related notes for PCS 211

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit