Answer:
Option b. 0.27 μC
Explanation:
By definition, Coulomb's law quantifies the amount of force between two stationary, electrically charged particles:
[tex]F= K\frac{q_{1}q_{2}}{d^{2} }[/tex] (1)
To calculate q₂, we have to find the distances between the charges:
[tex]d=\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}[/tex] (2)
[tex]d=\sqrt{(1-1)^{2} + (3-1)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2} + (2)^{2}}[/tex]
[tex]d= 2 mm}[/tex]
And now, we need to calculate the electrostatic force, by using the relation between force and electric potential:
[tex]F= \frac{U(x)}{d}[/tex] (3)
[tex]F= \frac{49J}{2*10^{-3}m}[/tex]
[tex]F= 24500 N}[/tex]
Finally, we calculate the magnitude of the charge on the second object, using the ecuation (1):
[tex]F= K\frac{q_{1}q_{2}}{d^{2} }[/tex]
[tex]q_{2}=\frac{F*d^{2}}{K*q_{1}}[/tex]
[tex]q_{2}=\frac{24500N*(2*10^{-3}m)^{2}}{8.99*10^{9}Nm^{2}C^{-2}*40*10^{-6}C }[/tex]
q₂ = 2.72 × 10⁻⁷C = 0.27 μC
So, the answer is the option b. q₂ = 0.27 μC
Have a nice day!
