The figure shows the electric potential V as | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

The electric potential at any point $(x, y, z)$ (all in meters) in space is given by $V = 5x^2$ volt. The electric field at the point $(1, 2, 3) \text{ m}$ is $\overrightarrow{E} = $ . . . . . . $\text{N/C}$.

The potential $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

The electric potential as a function of $x, y$ is given by $V = 5(x^2 - y^2) \text{ V}$. The electric field at a point $(2, 3) \text{ m}$ is . . . . . . $\text{V/m}$.

If $V$ is the electric potential at a given point,then the electric field $E_x$ in the $x$-direction at that point is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module