JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 17)
A point charge $$2\times10^{-2}~\mathrm{C}$$ is moved from P to S in a uniform electric field of $$30~\mathrm{NC^{-1}}$$ directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m respectively, the work done by electric field will be
600 mJ
$$-1200$$ mJ
1200 mJ
$$-600$$ mJ
Explanation
_29th_January_Evening_Shift_en_17_1.png)
$$w = \int {F\,.\,ds} $$
$$\overrightarrow F = q\overrightarrow E = 2 \times {10^{ - 2}} \times 30\widehat i = 0.6N\widehat i$$
$$w = \overrightarrow F .\overrightarrow d = (0.6\widehat i)\,.\,( - \widehat i, - 2\widehat j)$$
$$ = - 0.6$$ J
$$ = - 600$$ mJ
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