The figure shows two equipotential lines in t | vedclass

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

Question

(C) The parallel equipotential lines represent a uniform electric field in the region.The electric field $\vec{E}$ is perpendicular to the equipotenti

Vedclass · Accepted Answer

$-100 \ V/m, +200 \ V/m$

Vedclass · Answer

(C) The parallel equipotential lines represent a uniform electric field in the region.The electric field $\vec{E}$ is perpendicular to the equipotential surfaces and points from higher potential to lower potential.The slope of the equipotential lines is $m = \frac{\Delta y}{\Delta x} = \frac{2-1}{6-4} = \frac{1}{2}$.Let $	heta$ be the angle the equipotential line makes with the $X$-axis. Then $	an 	heta = \frac{1}{2}$.This implies $\sin 	heta = \frac{1}{\sqrt{5}}$ and $\cos 	heta = \frac{2}{\sqrt{5}}$.The perpendicular distance $d$ between the two equipotential lines (potential difference $\Delta V = 4 \ V - 2 \ V = 2 \ V$) is calculated using the geometry. The horizontal distance between the lines at a constant $Y$ is $\Delta x = 6 - 4 = 2 \ cm = 0.02 \ m$.Thus,$d = \Delta x \sin 	heta = 0.02 	imes \frac{1}{\sqrt{5}} = \frac{0.02}{\sqrt{5}} \ m$.Using $\Delta V = E d$,we get $2 = E 	imes \frac{0.02}{\sqrt{5}}$,which gives $E = \frac{2 \sqrt{5}}{0.02} = 100 \sqrt{5} \ V/m$.The electric field vector points from $4 \ V$ to $2 \ V$ (towards the origin). The angle $\alpha$ that the electric field vector makes with the negative $X$-axis is $	heta$. Thus,$E_x = -E \sin 	heta = -(100 \sqrt{5}) 	imes \frac{1}{\sqrt{5}} = -100 \ V/m$.And $E_y = E \cos 	heta = (100 \sqrt{5}) 	imes \frac{2}{\sqrt{5}} = 200 \ V/m$.Therefore,$E_x = -100 \ V/m$ and $E_y = 200 \ V/m$. The correct option is $(C)$.

The figure shows two equipotential lines in the $XY$ plane for an electric field. The scale is shown. The $X$-component $E_x$ and $Y$-component $E_y$ of the electric field in the space between the equipotential lines are respectively:

Similar Questions

The figure shows a family of parallel equipotential surfaces and four paths along which an electron is made to move from one surface to another.
$(I)$ What is the direction of the electric field?
$(II)$ Rank the paths according to the magnitude of work done,greatest first.

Work done in moving a positive charge on an equipotential surface is

Electric field is always ...... to the equipotential surface at every point. (Fill in the gap)

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

What is an equipotential surface? Draw equipotential surfaces for:
$(1)$ $A$ single point charge
$(2)$ $A$ dipole (charges $+q$ and $-q$ at a small distance)
$(3)$ Two $+q$ charges at a small distance
$(4)$ $A$ uniform electric field.

Vedclass Test Series

Exam Paper Generator

Online Exam Module