Question #271642

A speck of dust has a mass of 1.0 x 10-18 kg and carries a charge equal to that of one electron.

Near to the Earth's surface it experiences a uniform downward electric field of strength

100 N C-1 and a uniform gravitational field of strength 9.8 N kg-1.

Draw a free-body force diagram for the speck of dust. Label the forces clearly.

Calculate the magnitude and direction of the resultant force on the speck of dust.


1
Expert's answer
2021-11-25T16:22:19-0500

Given:

m=1.0×1018kgm=1.0\times 10^{-18}\:\rm kg

q=1.6×1019Cq=1.6\times 10^{-19}\:\rm C

E=100N/CE=100\:\rm N/C

g=9.8N/kgg=9.8\:\rm N/kg





The resultant force


Fnet=mgqE=1.0×1018kg9.8N/kg1.6×1019C100N/C=6.2×1018NF_{\rm net}=mg-qE\\ =1.0\times 10^{-18}\:{\rm kg*9.8\:N/kg}\\-1.6\times 10^{-19}\:\rm C*100\:\rm N/C=-6.2\times 10^{-18}\:N

The net force directed upward.

The magnitude 6.2×1018N.6.2\times 10^{-18}\:\rm N.


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