Question

0.01 m 5. In an oil-drop experiment, two parallel conducting plates are connected to a power supply with a constant voltage of 100 V. The separation between the plates is 0.01 m. A 4.8x10-6 kg oil drop is suspended in the region between the plates. Use g = 10 m/s. a. What is the direction of the electric field between the plates? b. What is the magnitude of the electric field between the plates? c. What is the sign and magnitude of the electric charge on the oil drop when it stays stationary? The mass of the drop is reduced to 3.2*10-16 kg because of vaporization. d. What is the acceleration of the drop?

Answer 1

$\begin{array}{l}given:\\voltage,\;V=100\;volt\\separation\;between\;plates,\;d=0.01\;m\\mass\;of\;oil\;drop,m=4.8\times10^{-16\;}\;kg\\part:a\\as\;the\;upper\;plate\;is\;connected\;to\;positive\;terminal\;and\;lower\;plate\;is\;connected\;to\\negative\;terminal.\;so\;upper\;plate\;will\;aquire\;positive\;charge\;and\;lower\;plate\;will\;aquire\;negative\;charge.\\as\;direction\;of\;electric\;field\;is\;always\;from\;positive\;charge\;to\;negative\;charge.\;thus\;here\;the\;\\electric\;field\;will\;be\boldsymbol\;\boldsymbol d\boldsymbol o\boldsymbol w\boldsymbol n\boldsymbol w\boldsymbol a\boldsymbol r\boldsymbol d\boldsymbol s\boldsymbol.\end{array}$

$\begin{array}{l}part:b\\potential,\;V=\frac1{4\pi\varepsilon_0}\frac qx=100\;volt\\electric\;field\;E=-\frac{dV}{dx}=-\frac d{dx}\left(\frac1{4\pi\varepsilon_0}\frac qx\right)=\frac1{4\pi\varepsilon_0}\frac q{x^2}=\frac Vx\\at\;midpoint\;x=\frac{0.01}2=0.005\;m\\so\;E=\frac{100}{0.005}=20000\;N/C\end{array}$

$\begin{array}{l}part:c\\for\;electric\;charge\;to\;stay\;stationary\\weight=electric\;force\;on\;charge\\as\;weight\;will\;act\;downward,\;so\;electric\;force\;should\;act\;upward\;to\;balance\;this\;force.\\mg=qE\\q=\frac{mg}E=\frac{4.8\times10^{-16}\times10}{20000}=2.4\times10^{-19}\;C\;\;\;\;\;\;\\and\;charge\;should\;be\;of\;negative\;sign,\;so\;that\;electric\;force\;will\;be\;upward.\\so\;q=-2.4\times10^{-19}\;C\;\;\end{array}$

$\begin{array}{l}part:d\\if\;mass\;of\;partice\;becomes,\;m=3.2\times10^{-16}\;kg\\u\sin g\;newtons\;2nd\;law\\net\;force\;=mass\times acceleration\\qE-mg=ma\\(2.4\times10^{-19}\times20000)-(3.2\times10^{-16}\;\times10)=3.2\times10^{-16}\;\times a\\a=\frac{1.6\times10^{-15}}{3.2\times10^{-16}}=5\;m/s^2\;\;\;\;\;(answer)\end{array}$