The electric potential at any point (x, y, z) | vedclass

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(D) The electric potential is given by $V = 3x^2$. The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -
abla V =

Vedclass · Accepted Answer

$-12$

Vedclass · Answer

(D) The electric potential is given by $V = 3x^2$. The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$. Calculating the partial derivatives: $\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3x^2) = 6x$. $\frac{\partial V}{\partial y} = 0$. $\frac{\partial V}{\partial z} = 0$. Thus,$\vec{E} = -6x \hat{i}$. At the point $(2, 0, 1)$,the $x$-coordinate is $2$. Substituting $x = 2$ into the expression for $\vec{E}$: $\vec{E} = -6(2) \hat{i} = -12 \hat{i} \ Vm^{-1}$. The magnitude of the electric field is $-12 \ Vm^{-1}$ along the $x$-direction.

The electric potential at any point $(x, y, z)$ in metres is given by $V = 3x^2$. The electric field at a point $(2, 0, 1)$ is (in $Vm^{-1}$)

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