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Q.

Figure shows the variation of electric field intensity E versus distance x. What is the potential difference between the points x=2m and at x=6m from O?

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a

60 V

b

40 V

c

30 V

d

80 V

answer is A.

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Detailed Solution

$E = - \frac{d V}{d t}$
$\Rightarrow \int_{V_{B}}^{V_{A}} d v = \int_{x = 2}^{x = 1} - \vec{E} . \vec{d} r$
$\begin{array}{l} = - \int_{x = 2}^{x = 4} 10 d x - \int_{x = 4}^{x = 6} (- 5 s + 30) d x \\ = - {[10 x]}_{2}^{4} - {[\frac{- 5 x^{2}}{2} + 30 x]}_{x = 4}^{6} \\ = - 10 (4 - 2) + \frac{5}{2} (36 - 16) - 30 (6 - 4) \\ = (- 20 + 50 - 60) V \\ = 30 V \end{array}$

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