Coulombs Law:
force=k q2 q1 / distance^2 solve for q2
a Negative Charge of -6.0 x 10^-6 C exerts an attractive force of 65 N on a second charge 0.050 m away. What is the magnitude on the second charge?
2 answers
Bc bobpursley annoyingly didnt put the answer and I do know the answer (I'm just being lazy looking up answers) here it is
Q1=(-6.0 x 10^-6) Q2=? F=65N d=0.050 m k=(9.0 x 10^-9)
65=(9.0x10^9)((-6.0x10^-6)Q2)/0.050^2
65 x(0.050^2)= 0.1625
0.1625 /(-6.0x10^-6)= (-2.71x10^4)
(-2.71x10^4)/(9.0x10^9)= (-3.0x10^-6)
ans=(-3.0x10^-6)
go thumbs down bobpursley's post
Q1=(-6.0 x 10^-6) Q2=? F=65N d=0.050 m k=(9.0 x 10^-9)
65=(9.0x10^9)((-6.0x10^-6)Q2)/0.050^2
65 x(0.050^2)= 0.1625
0.1625 /(-6.0x10^-6)= (-2.71x10^4)
(-2.71x10^4)/(9.0x10^9)= (-3.0x10^-6)
ans=(-3.0x10^-6)
go thumbs down bobpursley's post
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