P23_056 - q 2 (where x > d ), ~ E 1 points...

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56. Let q 1 = 4 Q< 0and q 2 =+ 2 Q> 0 (where we make the assumption that Q> 0). Also, let d =2 . 00 m, the distance that separates the charges. The individual magnitudes ¯ ¯ ¯ ~ E 1 ¯ ¯ ¯ and ¯ ¯ ¯ ~ E 2 ¯ ¯ ¯ are fgured From Eq. 23-3, where the absolute value signs For q 2 are unnecessary since this charge is positive. Whether we add the magnitudes or subtract them depends on iF ~ E 1 is in the same, or opposite, direction as ~ E 2 .A tp o i n t s l e F to F q 1 (on the x axis) the felds point in opposite directions, but there is no possibility oF cancellation (zero net feld) since ¯ ¯ ¯ ~ E 1 ¯ ¯ ¯ is everywhere bigger than ¯ ¯
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Unformatted text preview: q 2 (where x &gt; d ), ~ E 1 points leFtward and ~ E 2 points rightward so the net feld in this range is ~ E net = ~ E 2 ~ E 1 in the i direction . Although | q 1 | &gt; q 2 there is the possibility oF ~ E net = 0 since these points are closer to q 2 than to q 1 . Thus, we look For the zero net feld point in the x &gt; d region: ~ E 1 = ~ E 2 1 4 | q 1 | x 2 = 1 4 q 2 ( x d ) 2 which leads to x d x = r q 2 | q 1 | = r 1 2 . ThereFore, x = d 2 2 1 = 6 . 8 m specifes the position where ~ E net = 0 ....
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