Question

Two particles, one with charge −2.93×10−6 C and the other with charge 4.39×10−6 C , are 0.0209 m apart. What is the magnitude of the force that one particle exerts on the other?

magnitude of force:

Two new particles with identical positive charge q3 are placed the same 0.0209 m apart. The force between them is measured to be the same as that between the original particles. What is q3 ?

Answer #1

*Given,*

*The charges, q _{1} = - 2.93 * 10^{-6} C and
q_{2} = 4.39 * 10^{-6} C*

*The distance, r = 0.0209 m*

*a)*

*The electrostatic force, F _{e} = ( k
q_{1}q_{2}) / r^{2}*

*F _{e} = ( 9 * 10^{9} * 2.93 *
10^{-6} * 4.39 * 10^{-6} ) /
(0.0209)^{2}*

^{ } = 265.022 N

*b)*

*Let the charge of a new particle be q _{3}*

*The distance is, r = 0.0209 m*

*F _{e} = ( k q_{1}q_{2}) /
r^{2} ( q_{1} = q_{2})*

*F _{e} = k (q_{3})^{2} /
r^{2}*

*(q _{3})^{2} = F_{e} r^{2} /
k*

*(q _{3})^{2} = 265.022 *
(0.0209)^{2} / 9 * 10^{9}*

*(q _{3})^{2} = 1.29 *
10^{-11} C*

*q _{3} = 3.59 * 10^{-6} C*

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