The electric field $\vec E$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $\theta = 60^o$ with respect to the field lines as shown in the figure. The potential difference between points $1$ and $2$ is

  • A
    $\frac {E}{d \cos 60^o}$
  • B
    $Ed \cos 60^o$
  • C
    $\frac {Ed}{\cos 60^o}$
  • D
    $\frac {E}{d} \cos 60^o$

Explore More

Similar Questions

Two plates are separated by a distance of $20 \, cm$. The potential difference between them is $10 \, V$. The electric field between the two plates is ...... $V m^{-1}$.

If the electric potential at any point $(x, y, z) \, m$ in space is given by $V = 3x^2$ volt,the electric field at the point $(1, 0, 3) \, m$ will be ............

For a given electric field $\vec{E} = 2x\hat{i} + 3y\hat{j}$,find the potential at $(X, Y)$ if the potential at the origin is $5\, V$.

The following figure shows the electric potential $V$ as a function of position $x$ across $4$ regions on the $x$-axis. Which of the following is correct for the electric field $E$ in these regions?

The electric potential $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ is:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo