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Electrostatics Example Problems Definition Electric Force Interaction between two charges Units N Vector / Point Charge Scalar Formula vector πΉ=π ππ ππ π2 Uniform Field Formula πΉ = πΈππ Ignore signs of qs. Draw a picture for direction. πΉ πΈ= ππ Ignore signs of qs. Draw a picture for direction. Electric Field πΉ πΈ= ππ Electric Potential Energy Energy that can be converted into kinetic Electric Potential π= ππ ππ N/C J J/C V vector scalar scalar πΈ=π ππ π2 ππ = π π=π ππ ππ π ππ π Notes Maintain signs of qs. βππ = βππ πΈβπ Ignore direction. βπ = βπΈβπ Maintain signs of qs. Ignore direction. Electric Force In the rectangle below, a charge is to be placed at the empty corner to make a net force on the charge at corner A point along the vertical direction. What charge (magnitude and sign) must be placed at the empty corner? ANSWER: -3.3 µC +3 µC 4d C B d A +3 µC D +3 µC +1µC Electric field ? d = 5cm B C A D d a) Find the magnitude & direction -2µC +1µC of the electric field at corner C of the square. ANS: 1.5 x 106 N/C at 45° b) What charge qC should be placed at corner C so the net electric field at the center of the square is zero? ANS: -2µC c) Determine the magnitude and direction of the force exerted by three charges at A, B, and D on the fourth charge at C. ANS: 3N at 225° Electric potential +1 µC A d = 5cm ? B d d C -2 µC D d +1 µC E a) What is the total potential at C? ANS: -1.8 x 105 J/C b) A fourth charge q is placed at B such that the net potential at C is zero. What is the magnitude & sign of q? ANS: +1 µC c) How much work is done to move q from ANS: +0.06 J infinity to B? Uniform electric field An electron with speed β vo = 1.8 x 106 m/s is traveling parallel to an electric field of magnitude E = 7.7 x 103 N/C. (me = 9.11 x 10-31 kg, e = 1.6 x 10-19C) a) How far will it travel before it stops? ANS: 1.2 mm b) How much time will + elapse before it returns to its starting point? ANS: 2.6 ns β β β β β β + + + β + + + A good challenge An electron is accelerated horizontally from rest in a TV picture tube by a potential difference of 25,000 V (energy per charge). It then passes between two horizontal plates 6.0cm long and 1.3cm apart that have a potential difference of 250 V. At what angle ΞΈ will the electron be traveling after it passes between the plates? β β β β β β β β β β v = 9.37 x 107m/s F = 3.07 x 10-15N a = 3.38 x 1015m/s2 β ANSWER: 1.32° + + + + + + + t = 6.40 x 10-10s vx = 9.37 x 107m/s vy = 2.16 x 106m/s + + ΞΈ