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

Question

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

Accepted Answer

The electric field $\mathbf{E}$ is related to the electric potential $V$ by the negative gradient relationship: $\mathbf{E} = -
abla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}ight)$ [NCERT Class 12, Sec 2.6].

Given $V = 6xy - y + 2yz$, we calculate the partial derivatives:
1.  $\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y$
2.  $\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z$
3.  $\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y$

At the point $(1, 1, 0)$, substituting $x=1, y=1, z=0$:
*   $E_x = -\left(\frac{\partial V}{\partial x}ight) = -(6(1)) = -6$
*   $E_y = -\left(\frac{\partial V}{\partial y}ight) = -(6(1) - 1 + 2(0)) = -(5) = -5$
*   $E_z = -\left(\frac{\partial V}{\partial z}ight) = -(2(1)) = -2$

Thus, $\mathbf{E} = -6\hat{i} - 5\hat{j} - 2\hat{k} = -(6\hat{i} + 5\hat{j} + 2\hat{k})$ N/C.

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