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Calculating Electric Fields

Calculating Electric Fields. The electric field will be directly proportional to the charge setting up the field and inversely proportional to the square of the distance between that charge and where you are measuring the strength of the electric field. In other words,

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Calculating Electric Fields

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  1. Calculating Electric Fields

  2. The electric field will be directly proportional to the charge setting up the field and inversely proportional to the square of the distance between that charge and where you are measuring the strength of the electric field. In other words, Electric field = (Coulomb’s constant) (Charge on object setting up the field) (distance away from the charge)2 E is in N/C K=9x109 Nm2 C 2 Q is in Coulombs Distance is in meters E = kq d2

  3. Problem 1: What is the electric field .2 meters away from a +4 Coulomb charge? E = (9x109)(4) .22 The direction of the electric field would be away from the +4C charge since the direction is always the direction a small positive test charge would move. E=9x1011 N/C

  4. If you have an electric field created by two charges: • Calculate the electric field due to each charge separately • Place arrows on each electric field strength you just calculated that indicate the direction the small positive test charge would move when influenced by that particular electric field • If the arrows point the same direction, add the two electric fields together. • If the arrows point opposite directions, subtract the two electric fields and the resulting electric field will have the direction of whichever of the original electric fields was stronger.

  5. Problem #2: Calculate the electric field at point P below. Charge A = 6C Charge B = 2C Distance from A to P = .4 meters Distance from P to B = .2 meters A .4 P .2 B 6C 2C A positive test charge placed at point P would move to the right, away from charge A 3.375x1011 A positive test charge placed at point P would move to the left, away from charge B 4.5x1011 • 4.5 x 1011 • 3.375 x 1011 • 1.125 x 1011 N/C • The resulting direction • would be to the left • since that electric field • was stronger EA= (9x109)(6) .42 E= 3.375x1011 N/C EB= (9x109)(2) .22 E=4.5x1011 N/C Since the arrows point opposite directions, subtract the two magnitudes to find the net electric field

  6. Problem #3: Calculate the electric field at point P directly above A where: A=2C B=10C Distance from A to P = 5 meters Distance from A to B = 12 meters P 5 A 12 B 2C 10C Use the Pythagorean Theorem to find the distance between P and B EB= (9x109)(10) 132 EB=5.33x108 N/C EA= (9x109)(2) 52 EA= 7.2 x 108 N/C

  7. If we place a positive test charge at P, the arrows for the electric fields will be: EA = 7.2 x 108 N/C EB = 5.33 x 108 N/C We need to find the angle by referring back to our original picture Place the arrows head to tail: 5.33x108 13 5 x 7.2 x 108 12 Draw the resultant arrow tanx = 5/12 X=22.6º

  8. 5.33x108 The 22.6º would be here 7.2x108 So the angle we need would be 90+22.6=112.6° Now we can use the Law of Cosines to find the resultant magnitude: X2=(7.2x108)2+(5.33x108)2-2(7.2x108)(5.33x108)(cos112.6) X=1.05 x 109

  9. We will use the Law of Sines to find the resultant direction: 5.33x108 sin112.6 = sin x 1.05x109 5.33x108 112.6 1.05x109 7.2x108 X=27.9° West of North x

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