The electrical potential function for an electrical field directed parallel to the x-axis is shown in Fig. The magnitude of the electric field in the x-direction in the interval 4≤x≤8 is :
A
2.5NC−1
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B
5NC−1
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C
−2.5NC−1
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D
−5NC−1
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Solution
The correct option is A5NC−1 We know that, →E=−dVdx^n Therefore for region 4≤x≤8 slope of potential curve is −5 and its negative is 5 . therefore electric field in that region is 5 N/C