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Two very large horizontal sheets are 4.25 cm4.25 \mathrm{~cm} apart and carry equal but opposite uniform surface charge densities of magnitude σ\sigma. You want to use these sheets to hold stationary in the region between them an oil droplet of mass 486μg486 \mu \mathrm{g} that carries an excess of five electrons. Assuming that the drop is in vacuum.(b) what should σ\sigma be?

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Answered 1 year ago
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(b) To hold the oil droplet between the two plates, the electric force must equal the weight of the droplet as next

qE=mgqE = mg

Where qq is the charge of the oil droplet and mm is the mass of the droplet. EE is the electric field produced between the two plates and it could be given by

E=σϵE = \dfrac{\sigma}{\epsilon_{\circ}}

Where ϵ\epsilon_{\circ} is the electric constant and equals 8.85×1012 C2/Nm28.85 \times 10^{-12} \mathrm{~C^{2}/N\cdot m^{2}} (See Appendix F). Now plug this expression of EE into equation (1) and solve for σ\sigma

qE=mg(5e)σϵ=mgσ=ϵmg5e(2)\begin{gathered} qE = mg \\ (5e) \dfrac{\sigma}{\epsilon_{\circ}} = mg \\ \sigma = \epsilon_{\circ}\dfrac{mg}{5e} \tag{2} \end{gathered}

Now we can plug our values for m,g,em, g, e and ϵ\epsilon_{\circ} into equation (2) to get σ\sigma

σ=ϵmg5e=(8.85×1012 C2/Nm2)(486×109kg)(9.8 m/s2)5(1.60×1019C)=52.7C/m2\begin{aligned} \sigma &= \epsilon_{\circ}\dfrac{mg}{5e} \\ & = (8.85 \times 10^{-12} \mathrm{~C^{2}/N\cdot m^{2}}) \dfrac{(486 \times 10^{-9} \,\text{kg})(9.8 \mathrm{~m/s^{2}})}{5 ( 1.60 \times 10^{-19}\,\text{C})}\\ &= \boxed{52.7 \mathrm{C/m^{2}}} \end{aligned}

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