The electric field in a region is given by- E - 4axy -z - - 2ax2 -z - ax2y-zk- - where a is a positive constant- The equation of an equipotential surface will be of the form

At equipotential surface, V= ∫E.dr=constant ∫ (4axy √(z)) dx + ∫ (2ax^2 √(z)) dy+∫(ax^2y√(z))dz=constant (2ax^2y √(z))+(2ax^2y √(z))+(2ax^2y √(z))=constant (6ax ...

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Standard XII

Physics

Infinite Sheet

The electric ...

Question

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

z = \frac{constant}{[x^{2} y^{2}]}

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z = \frac{constant}{[x y^{2}]}

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z = \frac{constant}{[x^{4} y^{2}]}

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Solution

The correct option is C $z = \frac{constant}{[x^{4} y^{2}]}$
At equipotential surface,
$V = - \int \to E . \to d r = constant$
$\int (4 a x y \sqrt{z}) d x + \int (2 a x^{2} \sqrt{z}) d y + \int (\frac{a x^{2} y}{\sqrt{z}}) d z = constant$
$(2 a x^{2} y \sqrt{z}) + (2 a x^{2} y \sqrt{z}) + (2 a x^{2} y \sqrt{z}) = constant$
$(6 a x^{2} y \sqrt{z}) = constant$

$z = \frac{const}{x^{4} y^{2}}$

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The electric field in a region is given by: →E=(4axy√z)^i+(2ax2√z)^j+(ax2y√z)^k , where a is a positive constant. The equation of an equipotential surface will be of the form

The correct option is C z=constant[x4y2]At equipotential surface, V=−∫→E.→dr=constant ∫(4axy√z)dx+∫(2ax2√z)dy+∫(ax2y√z)dz=constant (2ax2y√z)+(2ax2y√z)+(2ax2y√z)=constant (6ax2y√z)=constant z=constx4y2

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