Physics Q34 – JEE-MAIN 2025

Home Physics Electric Charges and FieldsQuestion #110

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^{\circ}

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

ϵ_{0} = 8.85 \times 10^{- 12} \frac{F}{m}

and acceleration due to gravity,

g = 10 \frac{m}{s^{2}}

)

Clue 1 of 3: Force Analysis

Identify all forces acting on the charged bob at equilibrium. These include gravitational force, electric force, and tension in the string.

🥷

Ninja Strategy•Order of Magnitude Check

Estimate the order of magnitude of the result by combining the powers of 10 from the given values ( $10^{- 12} \times 10^{- 4} \times 10 / 10^{- 5} = 10^{- 10}$ ), which quickly narrows down the options to those around $10^{- 9}$ or $10^{- 10}$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Forces and Equilibrium Condition✦ Active

The bob is in equilibrium under three forces: gravitational force ( $m g$ ) acting downwards, electric force ( $F_{e}$ ) acting horizontally away from the sheet (since both sheet and bob have positive charge), and tension ( $T$ ) along the string. The string makes an angle of $45^{\circ}$ with the sheet, which means it makes an angle $θ = 90^{\circ} - 45^{\circ} = 45^{\circ}$ with the vertical. For equilibrium, the sum of forces in horizontal and vertical directions must be zero.

T \cos θ = m g (Vertical equilibrium)

T \sin θ = F_{e} (Horizontal equilibrium)

\tan θ = \frac{F_{e}}{m g}

Step 2: Calculate Electric Force and Substitute○ Expand

The electric field due to an infinite non-conducting sheet is $E = \frac{σ}{2 ϵ_{0}}$ . The electric force on the charge $q$ is $F_{e} = q E = \frac{q σ}{2 ϵ_{0}}$ . Substitute $F_{e}$ into the equilibrium equation and rearrange to solve for $σ$ .

\tan θ = \frac{q σ}{2 ϵ_{0} m g}

σ = \frac{2 ϵ_{0} m g \tan θ}{q}

Step 3: Substitute Values and Calculate○ Expand

Given values are $m = 100 mg = 10^{- 4} kg$ , $q = 10 μ C = 10^{- 5} C$ , $ϵ_{0} = 8.85 \times 10^{- 12} F/m$ , $g = 10 {m/s}^{2}$ , and $θ = 45^{\circ}$ . Note that $\tan (45^{\circ}) = 1$ .

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

σ = \frac{17.7 \times 10^{- 15}}{10^{- 5}} = 17.7 \times 10^{- 10} {C/m}^{2}

σ = 1.77 \times 10^{- 9} {C/m}^{2} = 1.77 {nC/m}^{2}

💡 Teacher's Secret Hint

Ensure correct unit conversions for mass and charge, and pay attention to powers of 10 in calculations.

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Physics Question 34 – JEE-MAIN 2025

Clue 1 of 3: Force Analysis

🧠 Full Solution Path