An electric field is given by vec{E} = e_1 ha | vedclass

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(A) The work done $W$ by an electric field on a charge $Q$ is given by the dot product of the force $\vec{F}$ and the displacement $\vec{r}$.Since $\v

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$Q(ae_1 + be_2)$

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(A) The work done $W$ by an electric field on a charge $Q$ is given by the dot product of the force $\vec{F}$ and the displacement $\vec{r}$.Since $\vec{F} = Q\vec{E}$,we have $W = \vec{F} \cdot \vec{r} = Q(\vec{E} \cdot \vec{r})$.Given $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$ and $\vec{r} = a \hat{i} + b \hat{j}$,the dot product is:$\vec{E} \cdot \vec{r} = (e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}) \cdot (a \hat{i} + b \hat{j} + 0 \hat{k}) = a e_1 + b e_2$.Therefore,the work done is $W = Q(ae_1 + be_2)$.

An electric field is given by $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$. $A$ charge $Q$ is moved by a displacement vector $\vec{r} = a \hat{i} + b \hat{j}$. The work done is:

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