Need help graphing this on a position time graph!!

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Ответ:

Dweath50

08.09.2020

Physics

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$ radial distance from the linear charge, $r=0.09\ m$ linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

We know that the electric field by the linear charge is given as:

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

Now the force on the given charge can be given as:

F=E.q

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

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