In Millikan\'s experiment, an oil drop f radius 1.97 mu m and density 0 873 g/cm

Question

In Millikan's experiment, an oil drop f radius 1.97 mu m and density 0 873 g/cm^3 is suspended in chamber C (see the figure) when a downward electric field of 1.56 times 10^5 N/C is applied. Find the charge on the drop, in terms of e. Number Units The equation E = 1/ 2pi elementof_0 qd/z^3 is approximation of the magnitude of the electric field of an electric dipole, at point axis. Consider a point P on that axis at distance z = 38.00 d from the dipole center (d is the separation distance between the particles of the dipole.) Let E_ be the magnitude of the field at point P as approximated by the equation below. Let E_ be the actual magnitude. What is the ration E_appr/E_act? Number units

Explanation / Answer

here,

initial speed , u = 53 km/s = 53000 m/s

time taken , t = 1.2 ns = 1.2 * 10^-9 s

the accelration , a = e * E /m

a = - 1.6 * 10^-19 * 49 /(9.1 * 10^-31)

a = - 8.62 * 10^12 m/s^2

a)

the final speed , v = u + a * t

v = 53000 - 8.62 * 10^12 * ( 1.2 * 10^-9) = 4.27 * 10^4 m/s

b)

the distance travelled , s = u * t + 0.5 * a * t^2

s = 53000 * 1.2 * 10^-9 - 0.5 *8.62 * 10^12 * (1.2 * 10^-9)^2

s = 5.7 * 10^-5 m

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