The electric field in a region is given by : E=(4axyz)i^+2a2zj^+ax2y/zk^ where a is a positive constant. The equation of an equipotential surface will be of the form

E=(4axyz)i^+2ax2zj^+ax2y/zk^V=−∫Exdx+∫Eydy+∫Ezdz⇒V=−2ax2y2+2ax2yz+2ax2yz+ constant. 3×2ax2yz= constant ⇒6ax2yz= constant ⇒z= constant x4y2

The electric field in a region is given by : E=(4axyz)i^+2a2zj^+ax2y/zk^ where a is a positive constant. The equation of an equipotential surface will be of the form

E=(4axyz)i^+2ax2zj^+ax2y/zk^V=−∫Exdx+∫Eydy+∫Ezdz⇒V=−2ax2y2+2ax2yz+2ax2yz+ constant. 3×2ax2yz= constant ⇒6ax2yz= constant ⇒z= constant x4y2

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The electric field in a region is given by : $E = (4 axy \sqrt{z}) \hat{i} + (2 a^{2} \sqrt{z}) \hat{j} + ({ax}^{2} y / \sqrt{z}) \hat{k}$ where a is a positive constant. The equation of an equipotential surface will be of the form

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z = constant / [x⁴ y²]

None

z = constant / [x³y²]

z = constant / [xy²]

answer is C.

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Detailed Solution

$\begin{array}{l} E = (4 axy \sqrt{z}) \hat{i} + (2 {ax}^{2} \sqrt{z}) \hat{j} + ({ax}^{2} y / \sqrt{z}) \hat{k} \\ V = - [\int E_{x} dx + \int E_{y} dy + \int E_{z} dz] \\ \Rightarrow V = - [2 {ax}^{2} y \sqrt{2} + 2 {ax}^{2} y \sqrt{z} + 2 {ax}^{2} y \sqrt{z}] + constant. \\ 3 \times 2 {ax}^{2} y \sqrt{z} = constant \\ \Rightarrow 6 {ax}^{2} y \sqrt{z} = constant \Rightarrow z = \frac{constant}{x^{4} y^{2}} \end{array}$