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NEET PHYSICSMedium

If potential (in volts) in a region is expressed as V(x,y,z) = 6xy - y + 2yz, the electric field (in N/C) at point (1,1,0) is

A

3î + 5ĵ + 3k̂

B

6î + 5ĵ + 2k̂

C

2î + 3ĵ + k̂

D

6î + 9ĵ + k̂

Step-by-Step Solution

The electric field E\mathbf{E} is related to the electric potential VV by the negative gradient relation: E=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\mathbf{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) [1].

Given V=6xy−y+2yzV = 6xy - y + 2yz, we calculate the partial derivatives:

  1. ∂V∂x=∂∂x(6xy−y+2yz)=6y\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y
  2. ∂V∂y=∂∂y(6xy−y+2yz)=6x−1+2z\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z
  3. ∂V∂z=∂∂z(6xy−y+2yz)=2y\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y

Substitute the coordinates of the point (1,1,0)(1, 1, 0) where x=1,y=1,z=0x=1, y=1, z=0:

  1. Ex=−(6(1))=−6E_x = - (6(1)) = -6
  2. Ey=−(6(1)−1+2(0))=−(6−1)=−5E_y = - (6(1) - 1 + 2(0)) = -(6 - 1) = -5
  3. Ez=−(2(1))=−2E_z = - (2(1)) = -2

Thus, E=−6i^−5j^−2k^=−(6i^+5j^+2k^)\mathbf{E} = -6\hat{i} - 5\hat{j} - 2\hat{k} = -(6\hat{i} + 5\hat{j} + 2\hat{k}). (Note: The provided options appear to list the vector with positive signs, likely due to a typo in the source or OCR, but the magnitude coefficients correspond to Option B).

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