The electric field in a region surrounding th | vedclass

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

Question

(A) The electric field $\vec{E}$ is uniform and directed along the positive $x$-axis.The relationship between electric field and electric potential is

Vedclass · Accepted Answer

$A$

Vedclass · Answer

(A) The electric field $\vec{E}$ is uniform and directed along the positive $x$-axis.The relationship between electric field and electric potential is given by $\vec{E} = -
abla V$,which implies that the electric potential $V$ decreases in the direction of the electric field.Since the electric field is directed towards the positive $x$-axis,the potential decreases as we move from left to right along the $x$-axis.Comparing the $x$-coordinates of the given points:Point $C$ is at $x = -a$Points $B$ and $D$ are at $x = 0$Point $A$ is at $x = a$Since the potential decreases as $x$ increases,the potential will be maximum at $C$ and minimum at $A$.

Similar Questions

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + y^2\hat j$. If the potential at $(0, 0, 0)$ is $2 \, V$,find the potential at $(1, 1, 1)$. (in $/3$)

The potential of the electric field produced by a point charge at any point $(x, y, z)$ is given by $V = 3x^2 + 5$,where $x, y$ are in metres and $V$ is in volts. The intensity of the electric field at $(-2, 1, 0)$ is

$A$ uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the potential $V$ is zero at $x = 0$,then its value at $x = +x$ will be:

The potential function of an electrostatic field is given by $V = 2x^2$. Determine the electric field strength at the point $(2 \ m, 0, 3 \ m)$.

If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module