The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of interaction between the two?
(k = 9.0 * 10© newton·metersÆ/coulombÆ)
A. 2.2 x10^11 newtons
B 4.1 x 10^-7 newtons
C. 5.0 x 10^11 newtons
D. 7.0 x 1-^-7 newtons

Question

The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of interaction between the two?
(k = 9.0 * 10© newton·metersÆ/coulombÆ)
A. 2.2 x10^11 newtons
B 4.1 x 10^-7 newtons
C. 5.0 x 10^11 newtons
D. 7.0 x 1-^-7 newtons

anonymous · Answer

@rvc

rvc · Answer

$$F=\frac{ 1 }{ 4 \pi \epsilon k }\frac{ Q_1O_2 }{ r^2 }$$

anonymous · Answer

i dont know how to plug that in.... @rvc

anonymous · Answer

i dont get coulombs law or whatever this is

rvc · Answer

Q_1=5C
Q2=7C
r=1.2m

anonymous · Answer

so would it be A? @rvc

anonymous · Answer

it is A, thank you that really helped! @rvc