For a given electric field vec{E} = 2xhat{i} | vedclass

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(A) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by $\Delta V = -\int \vec{E} \cdot d\vec{r}$.Given $\vec{E}

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$-X^2 - \frac{3}{2}Y^2 + 5$

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(A) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by $\Delta V = -\int \vec{E} \cdot d\vec{r}$.Given $\vec{E} = 2x\hat{i} + 3y\hat{j}$ and $V(0,0) = 5\, V$.Integrating from origin $(0,0)$ to point $(X, Y)$:$V(X, Y) - V(0,0) = -\int_{(0,0)}^{(X, Y)} (2x\hat{i} + 3y\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$$V(X, Y) - 5 = -\int_{0}^{X} 2x\, dx - \int_{0}^{Y} 3y\, dy$$V(X, Y) - 5 = -[x^2]_{0}^{X} - [\frac{3}{2}y^2]_{0}^{Y}$$V(X, Y) - 5 = -X^2 - \frac{3}{2}Y^2$$V(X, Y) = -X^2 - \frac{3}{2}Y^2 + 5$.

For a given electric field $\vec{E} = 2x\hat{i} + 3y\hat{j}$,find the potential at $(X, Y)$ if the potential at the origin is $5\, V$.

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