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(D) The work done by an electric field is independent of the path taken because the electric field is a conservative field. The work done $W$ is given

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$16$

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(D) The work done by an electric field is independent of the path taken because the electric field is a conservative field. The work done $W$ is given by $W = q \Delta V = -q \int_{A}^{B} \vec{E} \cdot d\vec{r} = -q \vec{E} \cdot \vec{d}$.First,calculate the displacement vector $\vec{d} = \vec{r}_B - \vec{r}_A = (2-1)\hat{i} + (1-2)\hat{j} + (3-0)\hat{k} = (1\hat{i} - 1\hat{j} + 3\hat{k}) \ m$.The electric field is $\vec{E} = (10\hat{i} + 30\hat{j}) \ Vm^{-1}$.The potential difference $\Delta V = V_B - V_A = -\vec{E} \cdot \vec{d} = -[(10)(1) + (30)(-1) + (0)(3)] = -[10 - 30] = 20 \ V$.The work done is $W = q \Delta V = (0.8 \ C)(20 \ V) = 16 \ J$.

$A$ region contains a uniform electric field $E = (10 \hat{i} + 30 \hat{j}) \ Vm^{-1}$. $A$ and $B$ are two points in the field at $(1, 2, 0) \ m$ and $(2, 1, 3) \ m$,respectively. The work done when a charge of $0.8 \ C$ moves from $A$ to $B$ in a parabolic path is (in $J$)

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