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Step-by-Step Solution
Step 1: Recall the relationship between electric field and potential
The electric field $\vec{E}$ at any point is related to the electric potential $V$ by the negative gradient of the potential:
$ \vec{E} = -\nabla V $
Step 2: Compute the gradient of the given potential
The potential is given by:
$ V(x, y, z) = 4x^2 $
The gradient operator in three dimensions is:
$ \nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}
$
Since $V$ depends only on $x$, the partial derivatives with respect to $y$ and $z$ are zero. Thus:
$ \frac{\partial V}{\partial x} = 8x, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0
$
Step 3: Find the electric field vector
Substitute these partial derivatives into the expression for the electric field:
$ \vec{E} = - \left( \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} \right) $
$ \vec{E} = - \left( \hat{i} \cdot 8x + \hat{j} \cdot 0 + \hat{k} \cdot 0 \right) = -8x \,\hat{i}
Step 4: Evaluate the electric field at the point (1, 0, 2)
At $x=1$, $y=0$, and $z=2$:
$ \vec{E}_{(1,0,2)} = -8 \times 1 \,\hat{i} = -8\,\hat{i}
Step 5: Interpret the result
The vector $ -8\,\hat{i} $ has a magnitude of 8 and points along the negative x-axis. Hence, the electric field at the point (1, 0, 2) is 8 volt/meter along the negative x-axis.
