In which region is the magnitude of the x-com | vedclass

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

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The relationship between the electric field $(E)$ and the potential $(V)$ is given by the formula $E_x = -\frac{dV}{dx}$.This means the magnitude of the $x$-component of the electric field,$|E_x|$,is equal to the absolute value of the slope of the $V-X$ graph,i.e.,$|E_x| = |\frac{dV}{dx}|$.To find the region where the magnitude of the electric field is maximum,we need to identify the region where the slope of the $V-X$ graph is steepest (i.e.,has the largest absolute value).Looking at the graph:- In regions $1$ and $4$,the potential is constant,so the slope is $0$,which means $E_x = 0$.- In regions $2$ and $3$,the potential changes with distance,so the slope is non-zero.- Comparing the steepness of the lines in regions $2$ and $3$,the line in region $2$ is much steeper than the line in region $3$.Therefore,the magnitude of the slope is maximum in region $2$,which implies the magnitude of the electric field is maximum in region $2$.

In which region is the magnitude of the $x$-component of the electric field maximum,if the potential $(V)$ versus distance $(X)$ graph is as shown?

Similar Questions

$A$ uniform electric field of $400 \, V/m$ is directed $45^\circ$ above the $x$-axis. The potential difference $V_A - V_B$ is: (Coordinates are in cm) (in $, V$)

If the electric potential in a region is given by $V = 4x^2$ volts,then the electric field at the point $(1, 0, 2) \ m$ is:

The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (4x^2 + 8x - 3) \ V$. The value of the electric field at $x = 0.5 \ m$,in $V/m$,is:

In a region,the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$,where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2 \ C$ situated at point $(1, 1, 1)$ is

If the potential energy of an electron and a proton at a distance $r$ is given by the formula $U = - \left( \frac{ke^2}{3r^3} \right)$,then which law of force applies?

Vedclass Test Series

Exam Paper Generator

Online Exam Module