JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Electrostatics

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 line charge towards 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 ______.

Answer: 6

Solution

The line charge produces a field

E_{line} = \frac{λ}{2 π ε _{0} r}

directed away from the line, while the sheet produces a uniform field

E_{sheet} = \frac{σ}{2 ε _{0}}

directed away from the sheet. For a point lying between the two, these contributions act in opposite directions.

At P, where

r_{P} = \frac{3}{π}

E_{P} = \frac{λ}{2 π ε _{0}} \cdot \frac{π}{3} - \frac{σ}{2 ε _{0}} = \frac{1}{2 ε _{0}} (\frac{λ}{3} - σ)

At Q, where

r_{Q} = \frac{4}{π}

E_{Q} = \frac{λ}{2 π ε _{0}} \cdot \frac{π}{4} - \frac{σ}{2 ε _{0}} = \frac{1}{2 ε _{0}} (\frac{λ}{4} - σ)

With

λ = 2 σ

the magnitudes become

E_{P} = \frac{σ}{6 ε _{0}}

and

E_{Q} = \frac{σ}{4 ε _{0}}

, so

\frac{E _{P}}{E _{Q}} = \frac{1/6}{1/4} = \frac{4}{6} .

Comparing with

\frac{E _{P}}{E _{Q}} = \frac{4}{a}

gives

a = 6

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer