A uniform electric field is prevailing in the | vedclass

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

Question

(B) The electric field $\vec{E}$ is uniform and directed along the $+X$-direction.Points $P(0,0)$ and $R(0,2)$ lie on the same plane perpendicular to

Vedclass · Accepted Answer

$V_P = V_R, V_P > V_Q$

Vedclass · Answer

(B) The electric field $\vec{E}$ is uniform and directed along the $+X$-direction.Points $P(0,0)$ and $R(0,2)$ lie on the same plane perpendicular to the electric field lines (the $YZ$-plane). Therefore,$P$ and $R$ lie on the same equipotential surface,which implies $V_P = V_R$.Electric potential decreases in the direction of the electric field. Since point $Q(2,0)$ is further along the $+X$-direction than point $P(0,0)$,the potential at $P$ must be greater than the potential at $Q$,i.e.,$V_P > V_Q$.Combining these,we get $V_P = V_R$ and $V_P > V_Q$.

$A$ uniform electric field is prevailing in the $X$-direction in a certain region. The coordinates of points $P$,$Q$,and $R$ are $(0,0)$,$(2,0)$,and $(0,2)$ respectively. Which of the following alternatives is true for the potentials at these points?

Similar Questions

At which point is the electric field maximum?

The work done to move a charge on an equipotential surface is

What is the direction of the lines of force at any point on the equipotential surface?

In an electric field due to charge $Q$,a charge $q$ moves from point $A$ to $B$ along an arc of a circle centered at $Q$,as shown in the figure. The work done is ($\varepsilon_0=$ permittivity of free space).

The angle between the electric lines of force and the equipotential surface is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module