Consider three point charges: Q_1=1.0 mu C, Q_2=-2.0 mu C, and Q_3=3.0 mu C. In
ID: 1586099 • Letter: C
Question
Consider three point charges: Q_1=1.0 mu C, Q_2=-2.0 mu C, and Q_3=3.0 mu C. In the following, all locations are on the x-axis. Vector quantities must be given with respect to the +x direction; i.e. F=-20 N means a force of 20 newtons in the -x direction. Answers must include proper units. Answer all questions. Q_1 is placed at x=0 and Q_2 at x=50 cm, Q_3 is very far away. What is the magnitude of the electric force on Q_2? What is the the electric field at x=25 cm? Q_1 and Q_2 are as before and Q_3 is added at x= 100 cm. What is now the Electric field at x=25 cm? A proton (m=1.67 Times 10^-27 kg) finds itself (it's been in therapy) in a uniform electric field of 2000V/m in the +x direction. If it is initially at rest, what will it's speed be after one second? What will its kinetic energy be in electron volts?Explanation / Answer
1.a) Force of Q3 on Q2 will be zero since it's placed at infinity. Force on Q2 will be only due to Q1.
F= K(Q1).(Q2)/(r^2) = 9*10^9*(1*10^-6)*(-2*10^-6)/.5^2= -.072 N
b) Electric field at a point = sum of the Electric fields due to each chrage = K(Q1)/d^2 +KQ2/d^2 = K(Q1 +Q2)/d^2 = -144000 V/m
2. Electric field at x=25cm is = K(Q1)/d^2 +KQ2/d^2 + K(Q3)/D^2 = K(Q1 +Q2)/d^2 + KQ3/D^2 , where d= 25cm , D= 75cm
thus, Electric field at x=25cm is = 12000 V/m
3. e= 1.6*10^-19 is the charge on proton
E= 2000V/m
acceleration = a = eE/m = 1.91*10^11 ms^-2
velocity= at= 1.91*10^11 m/s
kinetic energy = mv^2/2 = 7.664*10^-5 J