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ch22_p80 - 80 Let q1 denote the charge at y = d and q2...

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80. Let q 1 denote the charge at y = d and q 2 denote the charge at y = – d . The individual magnitudes G E 1 and G E 2 are figured from Eq. 22-3, where the absolute value signs for q are unnecessary since these charges are both positive. The distance from q 1 to a point on the x axis is the same as the distance from q 2 to a point on the x axis: r x d = + 2 2 . By symmetry, the y component of the net field along the x axis is zero. The x component of the net field, evaluated at points on the positive x axis, is E q x d x x d x = F H G I K J + F H G I K J + F H G I K J 2 1 4 0 2 2 2 2 π ε where the last factor is cos θ = x / r with θ being the angle for each individual field as measured from the x axis. (a) If we simplify the above expression, and plug in
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