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The electric potential at a point (x,y,z)(x, y, z) is given by V=−x2y−xz3+4V = -x^2y - xz^3 + 4. The electric field E⃗\vec{E} at that point is:

A

E⃗=(2xy+z3)i^+x2j^+3xz2k^\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}

B

E⃗=2xyi^+(x2+y2)j^+(3xz−y2)k^\vec{E} = 2xy\hat{i} + (x^2 + y^2)\hat{j} + (3xz - y^2)\hat{k}

C

E⃗=z3i^+xyzj^+z2k^\vec{E} = z^3\hat{i} + xyz\hat{j} + z^2\hat{k}

D

E⃗=(2xy−z3)i^+xy2j^+3z2xk^\vec{E} = (2xy - z^3)\hat{i} + xy^2\hat{j} + 3z^2x\hat{k}

Step-by-Step Solution

The electric field E⃗\vec{E} and electric potential VV are related by the gradient operation: E⃗=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right) [NCERT Class 12, Sec 2.6.1].

Given the potential V=−x2y−xz3+4V = -x^2y - xz^3 + 4 (assuming the trailing '+' implies a constant term like 4 from the standard examination question context):

  1. Calculate xx-component (ExE_x): ∂V∂x=∂∂x(−x2y−xz3+4)=−2xy−z3\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(-x^2y - xz^3 + 4) = -2xy - z^3 Ex=−∂V∂x=−(−2xy−z3)=2xy+z3E_x = -\frac{\partial V}{\partial x} = -(-2xy - z^3) = 2xy + z^3

  2. Calculate yy-component (EyE_y): ∂V∂y=∂∂y(−x2y−xz3+4)=−x2\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(-x^2y - xz^3 + 4) = -x^2 Ey=−∂V∂y=−(−x2)=x2E_y = -\frac{\partial V}{\partial y} = -(-x^2) = x^2

  3. Calculate zz-component (EzE_z): ∂V∂z=∂∂z(−x2y−xz3+4)=−3xz2\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(-x^2y - xz^3 + 4) = -3xz^2 Ez=−∂V∂z=−(−3xz2)=3xz2E_z = -\frac{\partial V}{\partial z} = -(-3xz^2) = 3xz^2

Combining these components, the electric field vector is E⃗=(2xy+z3)i^+x2j^+3xz2k^\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}.

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