Electrostatics Question 786

Question: The electric potential at a point (x, y, z) is given by V=x2yxz3+4. The electric field E at that point is

Options:

A) E=i^2xy+j^(x2+y2)+k^(3xzy2)

B) E=i^z3+j^xyz+k^z2

C) E=i^(2xyz3)+j^xy2+k^3z2x

D) E=i^(2xy+z3)+j^x2+k^3xz2

Show Answer

Answer:

Correct Answer: D

Solution:

[d] The electric field at a point is equal to negative of potential gradient at that point.

E=Vr=[Vxi^Vyj^Vzk^]

=[(2xy+z3)i^+j^x2+k^3xz2]



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