Q168 / 180 · COMEDK 2024 (Morning Shift) · Physics
If potential (in volt) in a region is expressed as $\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{xy}-\mathrm{y}+2 \mathrm{yz}$, the electric field (in $N C^{-1}$) at point $(1,0,1)$ is
- A$+7 \mathrm{j}$
- B$\mathrm{-6 i+7 j}$
- Correct answerC$\mathrm{-7 j}$
- D$\mathrm{6 i-7 j}$
Solution
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \dfrac{\partial V}{\partial x} \hat{i} + \dfrac{\partial V}{\partial y} \hat{j} + \dfrac{\partial V}{\partial z} \hat{k} \right)$.
Given $V(x, y, z) = 6xy - y + 2yz$.
Calculating the partial derivatives:
$\dfrac{\partial V}{\partial x} = \dfrac{\partial}{\partial x}(6xy - y + 2yz) = 6y$
$\dfrac{\partial V}{\partial y} = \dfrac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z$
$\dfrac{\partial V}{\partial z} = \dfrac{\partial}{\partial z}(6xy - y + 2yz) = 2y$
At the point $(1, 0, 1)$:
$\dfrac{\partial V}{\partial x} = 6(0) = 0$
$\dfrac{\partial V}{\partial y} = 6(1) - 1 + 2(1) = 6 - 1 + 2 = 7$
$\dfrac{\partial V}{\partial z} = 2(0) = 0$
Substituting these values into the expression for $\vec{E}$:
$\vec{E} = -(0 \hat{i} + 7 \hat{j} + 0 \hat{k}) = -7 \hat{j}$.
Answer: $\mathrm{-7 j}$
Given $V(x, y, z) = 6xy - y + 2yz$.
Calculating the partial derivatives:
$\dfrac{\partial V}{\partial x} = \dfrac{\partial}{\partial x}(6xy - y + 2yz) = 6y$
$\dfrac{\partial V}{\partial y} = \dfrac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z$
$\dfrac{\partial V}{\partial z} = \dfrac{\partial}{\partial z}(6xy - y + 2yz) = 2y$
At the point $(1, 0, 1)$:
$\dfrac{\partial V}{\partial x} = 6(0) = 0$
$\dfrac{\partial V}{\partial y} = 6(1) - 1 + 2(1) = 6 - 1 + 2 = 7$
$\dfrac{\partial V}{\partial z} = 2(0) = 0$
Substituting these values into the expression for $\vec{E}$:
$\vec{E} = -(0 \hat{i} + 7 \hat{j} + 0 \hat{k}) = -7 \hat{j}$.
Answer: $\mathrm{-7 j}$
Hence, the correct answer is option: 'C'.