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52. Let q 1 denote the charge at y = d and q 2 denote the charge at y = − d . The individual magnitudes ¯ ¯ ¯ ~ E 1 ¯ ¯ ¯ and ¯ ¯ ¯ ~ E 2 ¯ ¯ ¯ are fgured From Eq. 23-3, 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 2 + d 2 . By symmetry, the y component oF the net feld along the x axis is zero. The x component oF the net feld, evaluated at points on the positive x axis, is E x =2 µ 1 4 πε 0 ¶µ q x 2 + d 2 x √ x 2 + d 2 ¶ where the last Factor is cos θ = x/r with θ being the angle For each individual feld as measured From the x axis. (a) IF we simpliFy the above expression, and plug in
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