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Step-by-Step Solution
Step 1: Identify the given quantities
• Charge moved, $q = 2\times10^{-2}\,\mathrm{C}$.
• Uniform electric field, $\vec{E} = 30\,\mathrm{N\,C^{-1}}\,\hat{i}$ (directed along the positive x-axis).
• Initial position, $P = (1,\,2,\,0)\,\mathrm{m}$.
• Final position, $S = (0,\,0,\,0)\,\mathrm{m}$.
Step 2: Determine the displacement vector
The displacement $\vec{d}$ is given by the final coordinates minus the initial coordinates:
$$
\vec{d} = (0 - 1)\,\hat{i} + (0 - 2)\,\hat{j} + (0 - 0)\,\hat{k} = -\hat{i} - 2\,\hat{j}.
$$
Step 3: Calculate the force on the charge
In a uniform electric field $\vec{E}$, the electrostatic force on a charge $q$ is
$$
\vec{F} = q\,\vec{E}.
$$
Here,
$$
\vec{F} = \bigl(2\times10^{-2}\,\mathrm{C}\bigr)\,(30\,\mathrm{N\,C^{-1}})\,\hat{i}
= 0.6\,\hat{i}\,\mathrm{N}.
$$
Step 4: Compute the work done by the electric field
The work done by the electric field when the charge moves through displacement $\vec{d}$ is
$$
W = \vec{F}\cdot \vec{d}.
$$
Substituting the vectors,
$$
W = (0.6\,\hat{i}) \,\cdot\,\bigl(-\hat{i} - 2\,\hat{j}\bigr)
= 0.6 \times (-1) + 0.6 \times 0
= -0.6\,\mathrm{J}.
$$
(Note that there is no contribution from the $-\hat{j}$ component because $\vec{F}$ has no $\hat{j}$ component.)
Step 5: Express the result in millijoules
Since $1\,\mathrm{J} = 1000\,\mathrm{mJ}$,
$$
-0.6\,\mathrm{J}
= -0.6 \times 1000\,\mathrm{mJ}
= -600\,\mathrm{mJ}.
$$
Step 6: Final Answer
The work done by the electric field in moving the charge from $P$ to $S$ is
$$
\boxed{-600\,\mathrm{mJ}.}
$$
