A thin infinite sheet charge and an infinite line charge of respective charge densities and are placed parallel at 5 m distance from each other. [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A thin infinite sheet charge and an infinite line charge of respective charge densities $+ σ$ and $+ λ$ are placed parallel at 5 m distance from each other. Points $' P'$ and $' Q'$ are at $\frac{3}{π} m$ and $\frac{4}{π} m$ perpendicular distance from the line charge towards the sheet charge, respectively. $' E_{P}'$ and $' E_{Q}'$ are the magnitudes of resultant electric field intensities at point $' P'$ and $' Q'$ , respectively. If $\frac{E_{P}}{E_{Q}} = \frac{4}{a}$ for $2$ $| σ |$ $=$ $| λ |$ , then the value of $a$ is ________. [2023]

Ans.
(6)

$E_{A} = \frac{λ}{2 π ε_{0} r_{A}} - \frac{σ}{2 ε_{0}} {r_{A} = \frac{3}{π}} = \frac{1}{2 ε_{0}} [\frac{λ}{3} - σ]$

$E_{B} = \frac{λ}{2 π ε_{0} r_{B}} - \frac{σ}{2 ε_{0}} {r_{B} = \frac{4}{π}} = \frac{1}{2 ε_{0}} [\frac{λ}{4} - σ]$

$\frac{E_{A}}{E_{B}} = \frac{4}{3} (\frac{λ - 3 σ}{λ - 4 σ}) = \frac{4}{3} [\frac{2 σ - 3 σ}{2 σ - 4 σ}] = \frac{4}{3} [\frac{- σ}{- 2 σ}] = \frac{4}{6}$

Want Us to Email you About Special Offers & Updates?