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In a region, the potential is represented by V(x,y,z)=6x−8xy−8y+6yzV(x, y, z) = 6x - 8xy - 8y + 6yz, where VV is in volts and x,y,zx, y, z are in meters. The electric force experienced by a charge of 22 coulomb situated at a point (1,1,1)(1, 1, 1) is:

A

65 N6\sqrt{5} \text{ N}

B

30 N30 \text{ N}

C

24 N24 \text{ N}

D

435 N4\sqrt{35} \text{ N}

Step-by-Step Solution

The electric field E\mathbf{E} is related to the electric potential VV by the negative gradient relationship: 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}) [NCERT Class 12, Sec 2.6.1].

  1. Calculate Electric Field Components: Given V=6x−8xy−8y+6yzV = 6x - 8xy - 8y + 6yz Ex=−∂V∂x=−(6−8y)E_x = -\frac{\partial V}{\partial x} = -(6 - 8y) Ey=−∂V∂y=−(−8x−8+6z)=8x+8−6zE_y = -\frac{\partial V}{\partial y} = -(-8x - 8 + 6z) = 8x + 8 - 6z
  • Ez=−∂V∂z=−(6y)=−6yE_z = -\frac{\partial V}{\partial z} = -(6y) = -6y
  1. Evaluate Field at point (1,1,1)(1, 1, 1): Substitute x=1,y=1,z=1x=1, y=1, z=1: Ex=−(6−8(1))=2E_x = -(6 - 8(1)) = 2 Ey=8(1)+8−6(1)=10E_y = 8(1) + 8 - 6(1) = 10
  • Ez=−6(1)=−6E_z = -6(1) = -6 So, E=2i^+10j^−6k^ N/C\mathbf{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} \text{ N/C}.

  1. Calculate Electric Force: The force on a charge qq is F=qE\mathbf{F} = q\mathbf{E} [NCERT Class 12, Sec 1.7]. Given q=2 Cq = 2 \text{ C}: F=2(2i^+10j^−6k^)=4i^+20j^−12k^ N\mathbf{F} = 2(2\hat{i} + 10\hat{j} - 6\hat{k}) = 4\hat{i} + 20\hat{j} - 12\hat{k} \text{ N}.

  2. Calculate Magnitude of Force: ∣F∣=42+202+(−12)2=16+400+144=560|\mathbf{F}| = \sqrt{4^2 + 20^2 + (-12)^2} = \sqrt{16 + 400 + 144} = \sqrt{560} ∣F∣=16×35=435 N|\mathbf{F}| = \sqrt{16 \times 35} = 4\sqrt{35} \text{ N}.

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