As shown in the figure below,a point charge q | vedclass

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(B) The electric field is uniform and directed along the positive $x$-axis,so $\vec{E} = E \hat{i}$.The electric force on the charge $q$ is $\vec{F} =

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$-q E a$

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(B) The electric field is uniform and directed along the positive $x$-axis,so $\vec{E} = E \hat{i}$.The electric force on the charge $q$ is $\vec{F} = q \vec{E} = q E \hat{i}$.Since the electric force is a conservative force,the work done depends only on the initial and final positions,not on the path taken.The initial position is $P(a, b, 0)$ and the final position is $S(0, 0, 0)$.The displacement vector is $\vec{d} = \vec{S} - \vec{P} = (0 - a) \hat{i} + (0 - b) \hat{j} + (0 - 0) \hat{k} = -a \hat{i} - b \hat{j}$.The work done by the electric field is $W = \vec{F} \cdot \vec{d}$.$W = (q E \hat{i}) \cdot (-a \hat{i} - b \hat{j}) = -q E a (\hat{i} \cdot \hat{i}) - q E b (\hat{i} \cdot \hat{j})$.Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$,we get $W = -q E a$.

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