The potential of the electric field produced | vedclass

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Question

(A) The electric field intensity $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V$. Since the potential $V$ dep

Vedclass · Accepted Answer

$+12 \ Vm^{-1}$

Vedclass · Answer

(A) The electric field intensity $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V$. Since the potential $V$ depends only on $x$,the electric field is given by $E_x = -\frac{dV}{dx}$. Given $V = 3x^2 + 5$. Calculating the derivative: $\frac{dV}{dx} = \frac{d}{dx}(3x^2 + 5) = 6x$. Therefore,$E_x = -6x$. At the point $(-2, 1, 0)$,the $x$-coordinate is $-2$. Substituting $x = -2$ into the expression for $E_x$: $E_x = -6(-2) = +12 \ Vm^{-1}$. Thus,the intensity of the electric field at $(-2, 1, 0)$ is $+12 \ Vm^{-1}$.

The potential of the electric field produced by a point charge at any point $(x, y, z)$ is given by $V = 3x^2 + 5$,where $x, y$ are in metres and $V$ is in volts. The intensity of the electric field at $(-2, 1, 0)$ is

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