Find the potential function V(x, y) of an electrostatic field E→=2axyi^+ax2−y2j^ where ‘a’ is a constant.

E→=2axyi^+ax2−y2j^V(x,y)−V0=−∫(x,y)(0,0) E→.dr→ =ax2y−ay33⇒V(x,y)=V0+ax2y−ay33

Find the potential function V(x, y) of an electrostatic field E→=2axyi^+ax2−y2j^ where ‘a’ is a constant.

E→=2axyi^+ax2−y2j^V(x,y)−V0=−∫(x,y)(0,0) E→.dr→ =ax2y−ay33⇒V(x,y)=V0+ax2y−ay33

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Find the potential function V(x, y) of an electrostatic field $\vec{E} = (2 axy) \hat{i} + a (x^{2} - y^{2}) \hat{j}$ where ‘a’ is a constant.

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$V_{0} + {ax}^{2} y - \frac{{ay}^{3}}{3}$

$V_{0} - {ax}^{2} y + \frac{{ay}^{3}}{3}$

$V_{0} - {axy}^{2} - \frac{{ay}^{2}}{3}$

$V_{0} + {axy}^{2} + \frac{{ay}^{3}}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{E} = (2 axy) \hat{i} + a (x^{2} - y^{2}) \hat{j} \\ V_{(x, y)} - V_{0} = - \int_{(x, y)}^{(0, 0)} \vec{E} . d \vec{r} = ({ax}^{2} y - \frac{{ay}^{3}}{3}) \\ \Rightarrow V_{(x, y)} = V_{0} + {ax}^{2} y - \frac{{ay}^{3}}{3} \end{array}$