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(C) The potential energy $U$ of a charge $q$ in an electric field $\vec{E}$ is given by $U = qV$,where $V$ is the electric potential.The change in pot

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Does not change

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(C) The potential energy $U$ of a charge $q$ in an electric field $\vec{E}$ is given by $U = qV$,where $V$ is the electric potential.The change in potential energy is given by $\Delta U = -q \int \vec{E} \cdot d\vec{r}$.Given $\vec{E} = 10 \hat{i} 	ext{ V/m}$ and the velocity $\vec{v} = -2 \hat{j} 	ext{ m/s}$,the displacement vector $d\vec{r}$ is along the $\hat{j}$ direction.Since the electric field is in the $\hat{i}$ direction and the displacement is in the $\hat{j}$ direction,the dot product $\vec{E} \cdot d\vec{r} = (10 \hat{i}) \cdot (dy \hat{j}) = 0$.Because the work done by the electric field is zero,the potential energy of the charge does not change.

In a region of space,suppose there exists a uniform electric field $\vec{E} = 10 \hat{i} \text{ V/m}$. If a positive charge moves with a velocity $\vec{v} = -2 \hat{j} \text{ m/s}$,its potential energy:

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