The figure shows the electric potential V as | vedclass

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(B) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$. This means the magnitude of the electric field is equ

Vedclass · Accepted Answer

${E_1} = {E_3} = {E_5} = 0$ and ${E_2} < {E_4}$

Vedclass · Answer

(B) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$. This means the magnitude of the electric field is equal to the slope of the $V-x$ graph.In regions $1$,$3$,and $5$,the potential $V$ is constant,so the slope $\frac{dV}{dx} = 0$. Thus,${E_1} = {E_3} = {E_5} = 0$.In regions $2$ and $4$,the potential changes linearly. The magnitude of the electric field is given by the slope of the lines in these regions.From the graph,the slope of the line in region $4$ is steeper than the slope of the line in region $2$. Therefore,the magnitude of the electric field in region $4$ is greater than in region $2$,i.e.,${E_2} Thus,the correct relation is ${E_1} = {E_3} = {E_5} = 0$ and ${E_2} < {E_4}$.

The figure shows the electric potential $V$ as a function of distance through five regions on the $x$-axis. Which of the following is true for the electric field $E$ in these regions?

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