The variation of electrostatic potential V al | vedclass

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(D) The relationship between electric field $E_x$ and potential $V$ is given by $E_x = -\frac{dV}{dx}$.This means the electric field is the negative o

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$x$-component at point $C$ is along the positive $x$-axis.

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(D) The relationship between electric field $E_x$ and potential $V$ is given by $E_x = -\frac{dV}{dx}$.This means the electric field is the negative of the slope of the $V-x$ graph.At point $A$, the slope $\frac{dV}{dx}$ is positive, so $E_x = -(	ext{positive}) = 	ext{negative}$. Thus, the electric field at $A$ is along the negative $x$-axis.At point $B$, the graph reaches its peak, so the slope $\frac{dV}{dx} = 0$, which means $E_x = 0$.At point $C$, the slope $\frac{dV}{dx}$ is negative, so $E_x = -(	ext{negative}) = 	ext{positive}$. Thus, the electric field at $C$ is along the positive $x$-axis.Therefore, the correct statement is that the $x$-component at point $C$ is along the positive $x$-axis.

The variation of electrostatic potential $V$ along the $x$-direction is shown in the graph. Which of the following statements about the electric field is correct?

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