Answer:
[tex]F_{10} = - \frac{k_{q1}* qo }{d_{1} ^{2} } j[/tex]
Explanation:
The net force ( F ) on particle 0 due to particle 1 can be expressed as
[tex]F_{10} = \frac{K_{q1}*qo }{r_{10} ^{3} } r_{10}[/tex]
k = coulomb constant
[tex]r_{10}[/tex] = distance between charge o and charge 1
to get the vector form ( attached below ) we substitute [tex]d_{1}[/tex] for [tex]r_{10}[/tex] and [tex]- d_{1} j[/tex] for [tex]r_{10}[/tex]
The net force on [tex]q_{0}[/tex] by [tex]q_{1}[/tex] is , [tex]F=-K\frac{q_{0}*q_{1}}{r^{2} }j[/tex]
The force between two charges is given by Coulomb force law.
[tex]F=K\frac{q_{0}*q_{1}}{r^{2} }[/tex]
Where K is Coulomb constant, [tex]K=9*10^{9}Nm^{2}/C^{2}[/tex]
and r is the distance between both charges.
Here given that, [tex]q_{0}[/tex] is located as (0, 0, 0) and [tex]q_{1}[/tex] is located at [tex](0,d_{1}, 0)[/tex]
The force on [tex]q_{0}[/tex] by [tex]q_{1}[/tex] in the negative y - axis.
So that , Direction of force is, [tex]-j[/tex]
Hence, the net force on [tex]q_{0}[/tex] by [tex]q_{1}[/tex] is , [tex]F=-K\frac{q_{0}*q_{1}}{r^{2} }j[/tex]
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