2.Two charges q1 and q2 are separated by a distance 0.02mm in the air. The force between them is 100N. What will be the force between the q1 and q2 place 0.06mm apart in the air.
1) ∣F∣=kq,q2d2=9∗109∗14∗10−9∗23∗10−90.062|F|= \frac{kq,q_2}{d^2}=\frac{9*10^9*14*10^{-9}*23*10^{-9}}{0.06^2}∣F∣=d2kq,q2=0.0629∗109∗14∗10−9∗23∗10−9
=9∗14∗23∗1090.062=\frac{9*14*23*10^9}{0.06^2}=0.0629∗14∗23∗109
=80.5000∗109N=80.5000*10^9N=80.5000∗109N
=8.05∗1014N=8.05*10^{14}N=8.05∗1014N
2) F1=kq1q2d2F_1= \frac{kq_1q_2}{d^2}F1=d2kq1q2
100=9∗109(q1,q2)(0.021000)2100= \frac{9*10^9(q_1,q_2)}{(\frac{0.02}{1000})^2}100=(10000.02)29∗109(q1,q2)
=∣q1q∣=444.4∗10−20=|q_1q_|=444.4*10^{-20}=∣q1q∣=444.4∗10−20
F2=9∗109∗444.4∗10−20(0.061000)2F_2=\frac{9*10^9*444.4*10^{-20}}{(\frac{0.06}{1000})^2}F2=(10000.06)29∗109∗444.4∗10−20
F2=11.11NF _2=11.11NF2=11.11N
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