The electric potential at a point (x, y, z) is given by V = - x2y - xz3 + 4 The electric field at that point is (a) E = i 2xy + j(x² + y²) + k(3xz - y²) (b) E = iz³ + jxyz + kz² (c) E = i(2xy - z³) + jxy² + k 3z²x (d) E = i(2xy + z³) + jx² + k 3xz²

Answer: (d). E = - V with V = -x²y - xz³ + 4. Ex = -(∂ V)/(∂ x) = 2xy + z³ Ey = -(∂ V)/(∂ y) = x², Ez = -(∂ V)/(∂ z) = 3xz² So the answer is (d) E = i(2xy + z³) + j x² + k 3xz².

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Q119NEET 2009 · AIPMTEasy

Electric Potential & Equipotentials

The electric potential at a point (x, y, z) is given by V = $-$ x2y $-$ xz3 + 4

The electric field at that point is

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Answer: (d)

$\vec E = -\nabla V$ with $V = -x^2y - xz^3 + 4$ .
$E_x = -\dfrac{\partial V}{\partial x} = 2xy + z^3$
$E_y = -\dfrac{\partial V}{\partial y} = x^2$ , $E_z = -\dfrac{\partial V}{\partial z} = 3xz^2$
So the answer is (d) $\vec E = \hat i(2xy + z^3) + \hat j x^2 + \hat k\,3xz^2$ .

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