## Related questions with answers

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

Solution

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

$qE = mg$

Where $q$ is the charge of the oil droplet and $m$ is the mass of the droplet. $E$ is the electric field produced between the two plates and it could be given by

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

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

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

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

$\begin{aligned} \sigma &= \epsilon_{\circ}\dfrac{mg}{5e} \\ & = (8.85 \times 10^{-12} \mathrm{~C^{2}/N\cdot m^{2}}) \dfrac{(486 \times 10^{-9} \,\text{kg})(9.8 \mathrm{~m/s^{2}})}{5 ( 1.60 \times 10^{-19}\,\text{C})}\\ &= \boxed{52.7 \mathrm{C/m^{2}}} \end{aligned}$

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