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The electric field in a region is given by E= (4axysqrt z)i + (ax2sqrt 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

atul shukla , 11 Years ago
Grade 12
anser 1 Answers
Sumit Majumdar

Last Activity: 10 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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