Question

the total electric force on charge 1 measured counter-clockwise from the +x axis, in degrees? DQuestion 10 10.5 pts Charge 1 of +4 micro-coulombs and a mass of 0.3 kg lies on a horizontal surface and charge 2 of -5 micro-coulombs is placed directly 0.46 meters above this charge on the surface. What is the normal force, in Newtons, acting on the charge on the surface?

Answer 1

Since charges are unlike ,the electrostatic force is attractive,pulling the charge on the surface upwards.

Given $q_{1}=4\mu C=4*10^{-6}C$

$q_{2}=-5\mu C=-5*10^{-6}C$

distance, $d=0.46m$

Use Formula ,$F=kq_{1}q_{2}/d^{2}$

$F=9*10^{9}*4*10^{-6}C*5*10^{-6}C/(0.46m)^{2}$

$F=9*10^{9}*4*10^{-6}*5*10^{-6}/0.46^{2}$

$F=0.18/0.46^{2}$

$F=0.851N$

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Normal Force =Weight of charge at surface(acting downwards) - Force acting upwards

Normal Force =mg - F

$N=0.3kg*9.81-0.851N$

ANSWER: $N=2.092N$

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