The electric potential V(x, y, z) for a plana | vedclass

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(A) The electric field $E_x$ is related to the potential $V$ by the relation $E_x = -\frac{dV}{dx}$.$1$. For $x < -d$,$V = 0$,so $E_x = -\frac{d(0)}{d

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(A) The electric field $E_x$ is related to the potential $V$ by the relation $E_x = -\frac{dV}{dx}$.$1$. For $x $2$. For $-d \le x $E_x = -\frac{d}{dx}[-V_0(1 + \frac{x}{d})^2] = V_0 \cdot 2(1 + \frac{x}{d}) \cdot \frac{1}{d} = \frac{2V_0}{d}(1 + \frac{x}{d})$.At $x = -d$,$E_x = 0$. At $x = 0$,$E_x = \frac{2V_0}{d}$.$3$. For $0 \le x $E_x = -\frac{d}{dx}[-V_0(1 + 2\frac{x}{d})] = V_0 \cdot \frac{2}{d} = \frac{2V_0}{d}$.This is a constant value.$4$. For $x \ge d$,$V = -3V_0$,so $E_x = -\frac{d(-3V_0)}{dx} = 0$.Comparing these results with the given options,the graph in option $A$ correctly represents $E_x$ as a function of $x$.

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