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 electrostatic potential inside a charged spherical ball is given by $V = ar^2 + b$,where $r$ is the distance from its centre and $a$ and $b$ are constants. The volume charge density of the ball is [$\varepsilon_0$ = permittivity of free space].

The electric potential inside a charged sphere is given by $\phi = ar^2 + b$,where $r$ is the distance from the center,and $a$ and $b$ are constants. The charge density inside the sphere is .......

Electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Then,the electric force acting on a $2 \, C$ point charge placed at the origin will be......$N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module