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The electric field in a region is given by E= (4axy sqrt z )i + (ax 2 sqrt z) j + (2ax 2 y/sqrt z) k , where a is a positive constant. The equqtion of anequipotential surface will be of the form: A) Z= constant/(x 3 y 2 ) B) B) Z= constant/(xy 2 ) C) C) Z= constant/(x 4 y 2 ) D) D) none


The electric field in a region is given by E= (4axy sqrt z )i + (ax2 sqrt z) j +  (2ax2y/sqrt z) k , where a is a positive constant. The equqtion of anequipotential surface will be of the form:


 A)     Z= constant/(x3y2)


B)        B) Z= constant/(xy2)


C)        C)   Z= constant/(x4y2)


D)           D)  none


Grade:12

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
The equipotential surface would be given by:
V=\int dV=-\int \left (E_{x}dx+E_{y}dy+E_{z}dz \right )=-\int \left ( 4axy{\sqrt{z}}dx+ax^{2}{\sqrt{z}}dy+\frac{2ax^{2}y}{{\sqrt{z}}}dz \right )=-\left ( 2ax^{2}y{\sqrt{z}}+ax^{2}y{\sqrt{z}}+4ax^{2}y{\sqrt{z}} \right )=-7ax^{2}y{\sqrt{z}}
For this surface to be equipotential, the potential is a constant, hence,z=\frac{V}{49x^{4}y^{2}a^{2}}
So the correct option is option (c0.
Regards
Sumit

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