Electric field in a region is given by vec{E} | vedclass

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(C) The electric field is given by $\vec{E} = 10x\hat{i} + 5y\hat{j}$.We know that the relation between electric field and potential is $\Delta V = -\

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$2000$

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(C) The electric field is given by $\vec{E} = 10x\hat{i} + 5y\hat{j}$.We know that the relation between electric field and potential is $\Delta V = -\int \vec{E} \cdot d\vec{r}$.Let $V_0$ be the potential at the origin $(0, 0)$ and $V_P$ be the potential at $(10, 20)$.$V_P - V_0 = -\int_{(0,0)}^{(10,20)} (10x\hat{i} + 5y\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$.$500 - V_0 = -\int_{0}^{10} 10x \ dx - \int_{0}^{20} 5y \ dy$.$500 - V_0 = -[5x^2]_0^{10} - [\frac{5y^2}{2}]_0^{20}$.$500 - V_0 = -[5(100) - 0] - [\frac{5(400)}{2} - 0]$.$500 - V_0 = -500 - 1000$.$500 - V_0 = -1500$.$V_0 = 500 + 1500 = 2000 \ V$.

Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10 \ V/m^2$ and $B = 5 \ V/m^2$. If the electric potential at a point $(10, 20)$ is $500 \ V$, then the electric potential at the origin is . . . . . . $V$.

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