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NEET PHYSICSMedium

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

A

−(3i^+5j^+3k^)-(3\hat{i} + 5\hat{j} + 3\hat{k})

B

−(6i^+5j^+2k^)-(6\hat{i} + 5\hat{j} + 2\hat{k})

C

−(2i^+3j^+k^)-(2\hat{i} + 3\hat{j} + \hat{k})

D

−(6i^+9j^+k^)-(6\hat{i} + 9\hat{j} + \hat{k})

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 = -\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].

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

At the point (1,1,0)(1, 1, 0), substituting x=1,y=1,z=0x=1, y=1, z=0: Ex=−(∂V∂x)=−(6(1))=−6E_x = -\left(\frac{\partial V}{\partial x}\right) = -(6(1)) = -6 Ey=−(∂V∂y)=−(6(1)−1+2(0))=−(5)=−5E_y = -\left(\frac{\partial V}{\partial y}\right) = -(6(1) - 1 + 2(0)) = -(5) = -5

  • Ez=−(∂V∂z)=−(2(1))=−2E_z = -\left(\frac{\partial V}{\partial z}\right) = -(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}) N/C.

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