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 and angle of 45 ° with the sheet at equilibrium the charge density of sheet will be. (Given varepsilon 0 =8.85 × 10 -12 frac F m and acceleration due to gravity, .g=10 frac m s 2 \

Physics · Electrostatics

JEE Main 2025 — 2 April, Morning Shift — Question 58

A small bob of mass 100 mg and charge $+10 \mu \mathrm{C}$ is connected to an insulating string of

length 1 m . It is brought near to an infinitely long non-conducting sheet of charge density ' $\sigma$ ' as

shown in figure. If string subtends and angle of $45^{\circ}$ with the sheet at equilibrium the charge

density of sheet will be. (Given $\varepsilon_{0}=8.85 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}$

and acceleration due to gravity, $\left.g=10 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right\}$

$T \cos 45=E q$

$T \sin 45=m g$

$E q=m g$

$E=\frac{m g}{q}=\frac{\sigma}{2 \varepsilon_{0}}$

$\sigma=\frac{2 \varepsilon_{0} m g}{q}$

$=1.77 \mathrm{nC} / \mathrm{m}^{2}$

Answer key and solution verified before publishing.

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