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NEET PHYSICSMedium

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

A

E = i(2xy + z³) + jx² + k3xz²

B

E = i2xy + j(x² + y²) + k(3xz - y²)

C

E = iz³ + jxyz + kz²

D

E = i(2xy - z³) + jxy² + k3z²x

Step-by-Step Solution

  1. Relationship between Field and Potential: The electric field E\mathbf{E} is related to the electric potential VV by the negative gradient: E=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\mathbf{E} = -\nabla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}) [1].
  2. Partial Derivative with respect to 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=−(−2xy−z3)=2xy+z3E_x = -(-2xy - z^3) = 2xy + z^3
  3. Partial Derivative with respect to 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=−(−x2)=x2E_y = -(-x^2) = x^2
  4. Partial Derivative with respect to 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=−(−3xz2)=3xz2E_z = -(-3xz^2) = 3xz^2
  5. Resulting Vector: E=(2xy+z3)i^+x2j^+3xz2k^\mathbf{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}
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