Calculate the ratio of the electric force to the gravitational force between two

Question

Calculate the ratio of the electric force to the gravitational force between two electrons. A 5-mC charge experiences a force of 2 N directed north. What is the electric field (magnitude and direction) at the location of the charge? Kind the force on a -20-mC charge that replaces the 5-mC charge. A - 15-m() charge experiences a force of 25 N directed west. What is the electric field (magnitude and direction) at the location of the charge? If this charge is removed and not replaced, what is the electric field at this location? What is the electric field at a distance of 3 cm from 4 mC of negative charge? What is the electric field 2 m from 8 C of positive charge? What is the electric field at a distance of 0.2 nm from a carbon nucleus containing six protons and six neutrons? What is the electric field 5.3 times 10^-11 m from a proton? What is the electric field midway between charges of 2 C and 6 C separated by 2 m?

Explanation / Answer

74) Given , charge =-15mc

Force,F=25N and Electric field=?

We have , E=F/q=25*1000/15=5000/3 N/C

Due to an isolated negative charge E is towards the charge.

76) Distance ,r=2m

Charge q=8C

Electric field E=9*10^9*q/r^2

E=8*9*10^9/4=18*10^9 N/C

78) Proton has a charge q=1.6*10^(-19) C

Distance r= 5.3*10^(-11) m

Electric field E=9*10^9*q/r^2

E=1.6*10^(-19)*9*10^9/28.09*10^(-22)=0.5*10^(12) N/C

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