Question #132407
Two particles, with identical positive charges and a separation of 2.83 × 10^-2 m, are released from rest. Immediately after the release, particle 1 has an acceleration a1 whose magnitude is 5.56 × 10^3 m/s2, while particle 2 has an acceleration a2 whose magnitude is 11.7 × 10^3 m/s2. Particle 1 has a mass of 8.95 × 10^-6 kg. Find (a) the charge on each particle and (b) the mass of particle 2.
1
Expert's answer
2020-09-17T14:30:28-0400

1.Coulombs  lawF=kq1q2r2;F1F2=0;m1a1=m2a2=kq1q2r2;q1=q2=q;m1a1=m2a2=kq2r28.95×106×5.56×103==4.253×106×11.7×103=kq2r2;49.762×103=49.762×103;q=rFk;q=2.83×10249.762×1039×109=6.654×108C2.F1F2=0;m1a1m2a2=0;(Newtons  second law)m1a1=m2a2;m2=m1a1a2m2=8.95×106×5.56×10311.7×103=4.253×106kgAnswer:1.q=6.654×108C2.m2=4.253×106kg1.Coulomb's \;law\\F=k\frac{q_1q_2}{r^2};F_1-F_2=0;m_1a_1=m_2a_2=k\frac{q_1q_2}{r^2};\\q_1=q_2=q;m_1a_1=m_2a_2=k\frac{q^2}{r^2}\\8.95 × 10^{-6} ×5.56 × 10^3=\\=4.253 × 10^{-6}×11.7 \times 10^3=k\frac{q^2}{r^2};\\49.762\times10^{-3}=49.762\times10^{-3};\\q=r\sqrt{\frac{F}{k}};\\q=2.83\times10^{-2}\sqrt{\frac{49.762\times10^{-3}}{9\times10^9}}=6.654\times10^{-8}C\\2.F_1-F_2=0;m_1a_1-m_2a_2=0;\\(Newton's \;second\ law)\\m_1a_1=m_2a_2;\\m_2=\frac{m_1a_1}{a_2}\\m_2=\frac{8.95 × 10^{-6} ×5.56 × 10^3}{11.7 \times 10^3}=4.253 × 10^{-6}kg\\Answer:\\1.q=6.654\times10^{-8}C\\2.m_2=4.253 × 10^{-6}kg


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