Question #282877

Suppose you have two small "point" objects separated by a distance of 1 cm. Each object has a diameter of 10 −3 cm. One object has an excess of 3 × 10 10 electrons and the other has an excess of 2 × 10 10electrons on it. What is the electrostatic force that they exert on each other?


1
Expert's answer
2022-01-02T18:26:02-0500

Gives

Separation distance (r)=1cm

each object diameter d=103cm<<1cmd=10^{-3}cm<<1cm

Object assume point partical

Charge q1=n1eq2=n2eq_1=n_1e\\q_2=n_2e

n1=3×1010electronn_1=3\times10^{10} electron

n2=2×1010electronn_2=2\times10^{10}electron


q1=3×1010×1.6×1019=4.8×109Cq_1=3\times10^{10}\times1.6\times10^{-19}=4.8\times10^{-9}C

q1=2×1010×1.6×1019=3.2×109Cq_1=2\times10^{10}\times1.6\times10^{-19}=3.2\times10^{-9}C

Now electrostatic force

F=kq1q2r2F=\frac{kq_1q_2}{r^2}

Put value

F=9×109×4.8×109×3.2×1090.012F=\frac{9\times10^{9}\times4.8\times10^{-9}\times3.2\times10^{-9}}{0.01^2}

F=1.3824×103NF=1.3824\times10^{-3}N


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