Electric Field

Do the math for each of the five points.

1. Remember that the imaginary charge use to test the field is positive. A positive test charge would be pushed to the right by the positive charge on the left and to the left by the positive charge on the right. The charges are of the same magnitude. The distances are identical. The magnitude of the force from either charge will have the same value. The imaginary test charge isn't going anywhere. The field at the first location is zero.

   E1 =	∑  kq r2
   E1 =	0 N/C

2. A test charge placed at the second position would be pushed to the right (0°) by either charge on the diagram. Two fields pointing in the same direction are added by simple addition.

   E2 = ∑  kq r2

   E2 =  (9 × 109 N m2/C2)(1 × 10−6 C) ⎛ ⎜ ⎝ 1  +  1 ⎞ ⎟ ⎠ (6 m)2 (2 m)2

   E2 = 2500 N/C at 0°

3. The reasoning in this location is similar to the previous one. The only thing that's changed is the direction. The two charges would push a positive test charge located at position 3 to the left (180°).

   E3 = ∑  kq r2

   E3 =  (9 × 109 N m2/C2)(1 × 10−6 C) ⎛ ⎜ ⎝ 1  +  1 ⎞ ⎟ ⎠ (4 m)2 (8 m)2

   E3 = 700 N/C at 180°

4. This point was chosen because the geometry is simple. Place an imaginary test charge at position 4. The the charge on the left will push it down and to the right (↘). The the charge on the right will push it down and to the left (↙). Put these two vectors together in standard position (↙↘) on the diagram. Can you see how they'll meet at a right angle? Two vectors of equal magnitude at right angles. That forms a special triangle with interior angles of 45°, 45°, 90° and sides in the ratio 1:1:√2. Compute the magnitude of the field from either charge and multiply by √2 to get the magnitude of the resultant. The direction of the result is down on the diagram (270° or −90°).

   E4 = ∑  kq r2

   E4 =  ⎛ ⎜ ⎝ (9 × 109 N m2/C2)(1 × 10−6 C) ⎞ ⎟ ⎠ √2 (2√2 m)2

   E4 = 1,590 N/C at 270°

5. This last location gets special commendation as a "Royal PITA" for the amount of work we are about to do. Start by computing the magnitude of the field at position 5 from each charge separately.

   Eleft =  kq r2

   Eleft =  (9 × 109 N m2/C2)(1 × 10−6 C) (3 m)2

   Eleft = 1,000 N/C

   Eright =  kq r2

   Eright =  (9 × 109 N m2/C2)(1 × 10−6 C) (5 m)2

   Eright = 360 N/C

   Break each vector up into its components and combine the components in parallel directions. This location was chosen because it is directly above the charge on the left and because it forms a 3:4:5 triangle with the charge on the right. This means the sines and cosines can be computed without having to find any angles and without using a calculator.

   Ex =	Eleft cos θ	+	Eright cos θ
   Ex =	(1,000 N/C)(0)	+	(360 N/C)(−45)
   Ex =	(0 N/C)	+	(−288 N/C)
   Ex =	−288 N/C

   Ey =	Eleft sin θ	+	Eright sin θ
   Ey =	(1,000 N/C)(1)	+	(360 N/C)(35)
   Ey =	(+1,000 N/C)	+	(+216 N/C)
   Ey =	+1,216 N/C

   Finish using pythagorean theorem for magnitude…

   E5 = √(Ex2 + Ey2) E5 = √[(288 N/C)2 + (1,216 N/C)2] E5 = 1,250 N/C

   and tangent for direction.

   tan θ5 =	Ey Ex
   tan θ5 =	+1,216 N/C −288 N/C
   θ5 =	103°

   A vector has both magnitude and direction.

   E₅ = 1,250 N/C at 103°

Record the values at as many symmetric locations as possible and sketch in the vectors.

1. Start with the equation for the field around a point-like charge.

   E =	kq
   	r2

   Solve for charge.

   q =	Er2
   	k

   Substitute values. Remember to use radius, not diameter.

   q =	(3 × 106 N/C)(0.15 m)2
   	9 × 109 N m2/C2

   Compute the answer.

   q = 7.5 × 10−6 C = 7.5 μC

2. One coulomb is really a large quantity of charge that would require a really large number of Van de Graaff generators.

   1 C	= 133,333 VdGs
   7.5 × 10−6 C

   This is about the same as the number of schools in the United States.

Answer it.