The coordinates of A, B, C and D are (a, b, 0 | vedclass

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(A) The work done in moving a charge $q$ in an electric field is given by $W = -q \vec{E} \cdot \vec{d}$,where $\vec{d}$ is the displacement vector.Th

Vedclass · Accepted Answer

$-qEa$

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(A) The work done in moving a charge $q$ in an electric field is given by $W = -q \vec{E} \cdot \vec{d}$,where $\vec{d}$ is the displacement vector.The coordinates of $A$ are $(a, b, 0)$ and the coordinates of $D$ are $(0, 0, 0)$.The displacement vector $\vec{d} = \vec{D} - \vec{A} = (0 - a)\hat{i} + (0 - b)\hat{j} + (0 - 0)\hat{k} = -a\hat{i} - b\hat{j}$.The electric field is $\vec{E} = E\hat{i}$.The work done $W = -q \vec{E} \cdot \vec{d} = -q (E\hat{i}) \cdot (-a\hat{i} - b\hat{j})$.$W = -qE(-a) = qEa$.Wait,the work done by the external agent to move the charge is $W_{ext} = q(V_D - V_A)$.Since $V = -Ex$,$V_D = -E(0) = 0$ and $V_A = -E(a) = -Ea$.$W_{ext} = q(0 - (-Ea)) = qEa$.However,if we consider the work done by the electric field,$W_{field} = -qEa$.Given the options,the standard convention for such problems is the work done by the electric field,which is $-qEa$.

The coordinates of $A, B, C$ and $D$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$ respectively. How much work is done in moving a charge $q$ from $A$ to $D$ in a uniform electric field $\vec{E}$ directed along the positive $x$-axis?

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