The potential function of an electrostatic fi | vedclass

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Question

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

Vedclass · Accepted Answer

$\vec E = -8\hat i \ N C^{-1}$

Vedclass · Answer

(D) The electric field $\vec E$ is related to the potential $V$ by the relation $\vec E = -
abla V$.Since $V = 2x^2$,the electric field is given by $\vec E = -\left( \frac{\partial V}{\partial x} \hat i + \frac{\partial V}{\partial y} \hat j + \frac{\partial V}{\partial z} \hat k ight)$.Calculating the partial derivatives: $\frac{\partial V}{\partial x} = 4x$,$\frac{\partial V}{\partial y} = 0$,and $\frac{\partial V}{\partial z} = 0$.Thus,$\vec E = -(4x) \hat i$.At the point $(2 \ m, 0, 3 \ m)$,the $x$-coordinate is $2 \ m$.Substituting $x = 2$,we get $\vec E = -(4 	imes 2) \hat i = -8 \hat i \ N C^{-1}$.

The potential function of an electrostatic field is given by $V = 2x^2$. Determine the electric field strength at the point $(2 \ m, 0, 3 \ m)$.

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