JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Electrostatics

A small bob of mass 100 mg and charge

+ 10 μ C

is connected to an insulating string of length 1 m. It is brought near to an infinitely long nonconducting sheet of charge density '

σ

' as shown in figure. If string subtends an angle of

4 5^{\circ}

with the sheet at equilibrium the charge density of sheet will be:
(Given,

ε_{0} = 8.85 \times 1 0^{- 12} \frac{F}{m}

and acceleration due to gravity,

g = 10 m / s^{2}

)

Answer: (D)

Solution

The bob is held by its weight, the string tension, and the horizontal force

q E

from the sheet, where

E = \frac{σ}{2 ϵ _{0}}

. With the string hanging at

4 5^{\circ}

the horizontal and vertical pulls are equal, so

q E = m g

, i.e.

\frac{q σ}{2 ϵ _{0}} = m g

. Hence

σ = \frac{2 ϵ _{0} m g}{q} = \frac{2 \times 8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 4} \times 10}{10 \times 1 0 ^{- 6}} = 1.77 \times 1 0^{- 9} C/ m^{2} = 1.77 nC/ m^{2}

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option