The potential phi(x, y) of an electrostatic f | vedclass

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Question

(B) The relation between the electric field $\vec{E}$ and the electric potential $\phi$ is given by $\vec{E} = -
abla \phi$. Thus,$d\phi = -\vec{E} \

Vedclass · Accepted Answer

$-axy + C$

Vedclass · Answer

(B) The relation between the electric field $\vec{E}$ and the electric potential $\phi$ is given by $\vec{E} = -
abla \phi$. Thus,$d\phi = -\vec{E} \cdot d\vec{r}$. Given $\vec{E} = a(y \hat{i} + x \hat{j})$ and $d\vec{r} = dx \hat{i} + dy \hat{j}$. Substituting these into the expression: $d\phi = -a(y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$ $d\phi = -a(y dx + x dy)$ We know that $d(xy) = y dx + x dy$. Therefore,$d\phi = -a d(xy)$. Integrating both sides,we get: $\phi = -a \int d(xy) = -axy + C$. Comparing this with the given options,option $B$ is the correct answer.

The potential $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

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