91Ó°ÊÓ

Charges$q_1-q_2=+3.20×{10}^{-19}C$ are on a y-axis at distance$d=17.0cm$ from the origin.

The charge $q_3=+6.40×{10}^{-19}C$is moved gradually along the x-axis from$X=0mtox=+5.0m$

Differentiating the net force equation from Coulomb's law for the particle, we can get the required values of the x. Again, using the values of x, we can get the minimum and maximum force.

Formula:

The magnitude of the electrostatic force between any two particles,

$F=K\frac{\left|q_1\right|\left|q_2\right|\cos \mathrm{θ}}{r^2}$ (1)

Ifis the angle between the force and the x-axis, the cosine angle,

$\cos \mathrm{θ}=\frac{x}{\sqrt{x^2+d^2}}$ (2)

We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of the force exerted by$q_{1}or{q}_{2}on{q}_3$. Since$e=+1.60×{10}^{-19}Cthen{q}_1=q_2=+2eand{a}_3=4.0e$ and we have the net force value of the third particle using equation (1) and equation (2) as given:

$$\begin{gathered} F_{net}=2F\cos \mathrm{θ} \\ =\frac{2\left(2e\right)\left(4e\right)}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(x^2+d^2\right)}\frac{x}{\sqrt{x^2+d^2}} \\ =\frac{4e^{2}x}{\mathrm{Ï€}{\mathrm{Î\mu }}_o{\left(x^2+d^2\right)}^{3/2}}.....................................\left(3\right) \end{gathered}$$

To find where the force is at an extremum, we can set the derivative of this expression equal to zero and solve for x, but it is good in any case to graph the function for a fuller understanding of its behavior, and as a quick way to see whether an extremum point is a maximum or a minimum. In this way, we find that the value coming from the derivative procedure is a maximum (and will be presented in part (b) and that the minimum is found at the lower limit of the interval. Thus, the net force is found to be zero at x=0 , which is the smallest value of the net force in the interval$5.0mâ‰\yen xâ‰\yen 0m.$

Hence, the value of x for the minimum is zero.

Similarly, using the concept of part (a) calculations, taking the derivative of equation (3) and equating it to zero, the maximum is found to be at:
$$\begin{gathered} x=\frac{d}{\sqrt{2}} \\ =\frac{17cm}{\sqrt{2}} \\ =12cm \end{gathered}$$

Hence, the value of x for the maximum value is 12 cm.

Substituting the value of x = 0 in equation (3), the value of the minimum net force is given as:$F_{net}=0N$

Hence, the value of minimum force is zero.

Substituting the value of$x=\frac{d}{\sqrt{2}}$ in equation (3), the value of the minimum net force is given as:

$$\begin{gathered} F_{net}=\frac{4e^2\left(12cm\right)}{\mathrm{Ï€}{\mathrm{Î\mu }}_0{\left({\left(12cm\right)}^2+{\left(17cm\right)}^2\right)}^{3/2}} \\ =+4.9×{10}^{-26}N. \end{gathered}$$

Hence, the value of maximum force is$4.9×{10}^{-26}N$ .