JEE Main 2025 Apr 2 Shift 1, Physics Q19: Coulomb's Law And Electric Field

JEE Main2025Apr 2, Shift 1Physics

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 non–conducting sheet of charge density $σ$ as shown in figure. If string subtends an angle of 45° 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, $g = 10 m/s^{2}$ )

A
$0.885 nC/m^{2}$
B
$17.7 nC/m^{2}$
C
$885 nC/m^{2}$
D
$1.77 nC/m^{2}$

Solution

The bob is in equilibrium under three forces: tension $T$ along the string, gravity $m g$ downward, and the horizontal electrostatic repulsion $F_{e} = q E$ from the sheet, where $E = \frac{σ}{2 ϵ _{0}}$ .

The string makes $45°$ with the (vertical) sheet, so it makes $45°$ with the vertical as well. Force balance:

$tan 45° = \frac{F _{e}}{m g} = \frac{q E}{m g} \Rightarrow q E = m g$

$q \cdot \frac{σ}{2 ϵ _{0}} = m g \Rightarrow σ = \frac{2 ϵ _{0} m g}{q}$

Substitute numerical values: $m = 100 mg = 1 0^{- 4} kg$ , $q = 10 μ C = 1 0^{- 5} C$ :

$σ = \frac{2 \times 8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 4} \times 10}{1 0 ^{- 5}}$

$σ = \frac{17.7 \times 1 0 ^{- 15}}{1 0 ^{- 5}} = 17.7 \times 1 0^{- 10} C/m^{2} = 1.77 nC/m^{2}$

$∴$ Option (4) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q19: Coulomb's Law And Electric Field

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