A small bob of mass 100mathrm~mg$100\mathrm{~mg}$ and charge +10mathrm~mu C$+10\mathrm{~\mu C}$ is connected to an insulating string of length 1mathrm~m$1\mathrm{~m}$. It is brought near to an infinitely long non-conducting sheet of charge density 'sigma$\sigma$' as shown in figure. If string subtends an angle of 45^circ$45^{circ}$ with the sheet at equilibrium the charge density of sheet will be: (Given, epsilon_0 = 8.85times 10^-12fracmathrmFmathrmm$\epsilon_0 = 8.85\times 10^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ and acceleration due to gravity, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s^2}$)

### Related Formula E = fracsigma2epsilon_0 quad text(field of infinite non-conducting charged sheet)$$E = \frac{\sigma}{2\epsilon_0} \quad \text{(field of infinite non-conducting charged sheet)}$$ tantheta = fracF_emg$$\tan\theta = \frac{F_e}{mg}$$ ### Core Logic In equilibrium, three forces act on the suspended charged bob: 1. Tension T$T$ directed along the string at theta = 45^circ$\theta = 45^{circ}$ with the vertical sheet. 2. Weight mg$mg$ directed vertically downwards. 3. Electrostatic repulsion force F_e = qE$F_e = qE$ acting horizontally away from the sheet. From the balance of forces in vertical and horizontal directions: T cos(45^circ) = mg$$T \cos(45^{circ}) = mg$$ T sin(45^circ) = q E$$T \sin(45^{circ}) = q E$$ Dividing the two equations: tan(45^circ) = fracq Emg = 1 implies q E = mg$$\tan(45^{circ}) = \frac{q E}{mg} = 1 \implies q E = mg$$ Substitute the expression for E$E$: q left(fracsigma2epsilon_0right) = mg implies sigma = frac2 epsilon_0 m gq$$q \left(\frac{\sigma}{2\epsilon_0}\right) = mg \implies \sigma = \frac{2 \epsilon_0 m g}{q}$$ Now plug in the given numerical values: - m = 100mathrm~mg = 100 times 10^-6mathrm~kg = 10^-4mathrm~kg$m = 100\mathrm{~mg} = 100 \times 10^{-6}\mathrm{~kg} = 10^{-4}\mathrm{~kg}$ - q = +10mathrm~mu C = 10 times 10^-6mathrm~C = 10^-5mathrm~C$q = +10\mathrm{~\mu C} = 10 \times 10^{-6}\mathrm{~C} = 10^{-5}\mathrm{~C}$ - g = 10mathrm~m/s^2$g = 10\mathrm{~m/s^2}$ - epsilon_0 = 8.85 times 10^-12mathrm~F/m$\epsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$ sigma = frac2 times (8.85 times 10^-12) times 10^-4 times 1010^-5$$\sigma = \frac{2 \times (8.85 \times 10^{-12}) \times 10^{-4} \times 10}{10^{-5}}$$ sigma = 17.7 times 10^-10mathrm~C/m^2 = 1.77 times 10^-9mathrm~C/m^2 = 1.77mathrm~nC/m^2$$\sigma = 17.7 \times 10^{-10}\mathrm{~C/m}^2 = 1.77 \times 10^{-9}\mathrm{~C/m}^2 = 1.77\mathrm{~nC/m}^2$$ ### Step 1: Final Conclusion The charge density of the sheet is 1.77mathrm~nC/m^2$1.77\mathrm{~nC/m}^2$. ### Pattern Recognition For a charge hanging near a vertical charged sheet, the equilibrium angle is governed by tantheta = fracF_emg$\tan\theta = \frac{F_e}{mg}$. For theta = 45^circ$\theta = 45^{circ}$, the horizontal force equals the vertical force (F_e = mg$F_e = mg$). Be careful to use the field of a non-conducting sheet: E = fracsigma2epsilon_0$E = \frac{\sigma}{2\epsilon_0}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics