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Question

(B) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.Given that the electric field

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$V_x = -x E_0$

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(B) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.Given that the electric field is uniform and directed along the positive $X$-axis,we have $E = E_0$.Substituting this into the formula: $E_0 = -\frac{dV}{dx}$.Rearranging the terms to integrate: $dV = -E_0 dx$.Integrating both sides from the reference point $x = 0$ (where $V = 0$) to a point $x$ (where potential is $V_x$):$\int_{0}^{V_x} dV = -\int_{0}^{x} E_0 dx$.$[V]_{0}^{V_x} = -E_0 [x]_{0}^{x}$.$V_x - 0 = -E_0 (x - 0)$.$V_x = -x E_0$.Therefore,the potential at $x$ is $-x E_0$.

$A$ uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the potential $V$ is zero at $x = 0$,then its value at $x = +x$ will be:

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