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Question

(B) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x}

Vedclass · Accepted Answer

$-2\hat{i} - 3\hat{j} + \hat{k}$

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(B) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Given $V = 2x + 3y - z$,we calculate the partial derivatives:$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(2x + 3y - z) = 2$$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(2x + 3y - z) = 3$$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(2x + 3y - z) = -1$Substituting these values into the expression for $\vec{E}$:$\vec{E} = -(2\hat{i} + 3\hat{j} - 1\hat{k})$$\vec{E} = -2\hat{i} - 3\hat{j} + \hat{k}$

The electric potential in a region is represented as $V = 2x + 3y - z$. The expression for the electric field strength is:

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