Question

2. -5.00 cm from rest in a constant electric field with magnitude 2.50 x103 N/C pointing A proton is released at in the positive y-direction. a. Determine the change in electrical potential energy of the proton when is it at x-2.60 cm. Did the proton gain or lose potential energy? b. Find the final speed of the proton at x 2.60 cm. [mass of proton 1.67 x 1027kg; charge of proton +1.60 x 1019C]

Answer 1

The lorce experien ce d by charged par ticle, in elechnic field is given by an The electic Potental energy chfferente between two points in amount of work dore by charge, without acceleration, trom the elechric feld is defined as an some exteynal foyce in moving point to the On e other. The amawnt ot woyk done by xternal an force inmoving a test chage 9. from A to B, opposite to electric field E is given by, WAF +0 A E.J, (becanse here, F= teal force .F-F .E If the electric held is unifom,ie does not de pend On xwe have, Uk-UA .E(-X,) AU 2 AU

Here i Au is paesi tive,the partcle will gain poten tial energy. 1f AU is ngative, the particle will lose potential energy Held is a onservative know, ekctric Nov Cor We Field. In a conseyvative field, the total energy of the Particle remains Constont ene7gy = conshant je, kinehc energy Potanhal Now, kinetic energy, k mv 2 Where masS velocity have Now we 1 L.E.XB - m (=X) 2. 24.E m 29.E if the particle star ts at rest hav, VA 0 from pont X, we

Now, lets solve he given question Here, we have E 2.5x10 N/C) i 0.05 m 1. (2.60 Cm) C0.0 26 m) - 2 = 1.Cox Jo17 -27 M1.67 X1o* kg a) Using , change in potendial energy -17 = (1.Cx 10) (2.5X1o N/e ?0.05m-0.0) Au -17 3.04 X10 the proton loses pdential negativre So AU is energy b) using the velocity of the proton at X 2.cm 2xCi.GX1oe)(1.5AO N) i(0.05 mi-0.026m2) (3.67X109) -27 3.64 X101 m/s Ve