Electric field solving
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Electric field solving

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    Electric field solving Electric field solving Presentation Transcript

    • Example continued Field due to q 1 E = 10 10 N.m 2 /C 2 10 X10 -9 C/(3m) 2 = 11 N/C in the y direction. Recall E =kq/r 2 and k=8.99 x 10 9 N.m 2 /C 2 E y = 11 N/C E x = 0 Field due to q 2 5 E = 10 10 N.m 2 /C 2 15 X10 -9 C/(5m) 2 = 6 N/C at some angle  Resolve into x and y components  E E y =E sin  C E x =E cos  C  Now add all components E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C Magnitude x y q 1 =10 nc q 2 =15 nc 4 3
    • Example continued E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C E Magnitude of electric field      tan -1 E y /E x = atan (14.6/-4.8)= -72.8 deg x q 1 =10 nc q 2 =15 nc 4 3
    • Motion of point charges in electric fields
      • When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newton’s Law:
      • a = F/m = qE/m for uniform electric fields
      • a = F/m = mg/m = g for uniform gravitational fields
      • If the field is uniform, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different.
      • Replace g by qE/m in all equations;
      • For example, In y =1/2at 2 we get y =1/2(qE/m)t 2