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JEE MAIN_2026
06-04-26_S1
Question
A three coulomb charge moves from the point $(0,-2,-5)$ to the point $(5,1,2)$ in an electric field expressed as $\vec{E}=2 x \hat{\mathrm{i}}+3 \mathrm{y}^2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}$. The work done in moving the charge is $\_\_\_\_$ J.
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Solution
$\begin{aligned} & \vec{F}=q \vec{E}=6 x \hat{\imath}+9 y^2 \hat{\theta}+4 \hat{k} \\ & W=\int F_x d x+\int F_y d y+\int F_z d z \\ & W=\int_0^5 6 x d x+\int_{-2}^1 9 y^2 d y+\int_{-5}^2 4 d z \\ & =186 \mathrm{~J}\end{aligned}$
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