Respuesta :
Answer:
The force on the 2.00-uC charge is [tex]36 \times 10^{10} \mathrm{N}[/tex]
Explanation:
We know that force between two charges is given by Coulomb’s law,
[tex]\mathrm{F}=\mathrm{k} \frac{q 1 q 2}{r * r}[/tex]
Where k = Coulomb’s constant =
[tex]9.0 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}[/tex]
And q1 and q2 are the charges given to be = -4.00-uC and 2.00-uC charges
And r = distance between the charges = 20 cm = 0.2 m
Substituting the given values in the formula we get force applied on [tex]2.00\ \mu C[/tex] charge,
F = [tex]36 \times 10^{10} \mathrm{N}[/tex] attractive force which is the required answer.
The force on the 2.00-uC charge is 1.8N towards the right.
Let us assume Q = -4.00μC and charge q = 2.00μC. Distance between them is r = 0cm = 0.2m
Coulomb's Law:
According to Coulomb's law, the electrostatic force between two charges Q and q separated by a distance r is given by:
[tex]F=k\frac{Qq}{r^2}[/tex]
where k is Coulomb's constant
[tex]F=\frac{9\times10^9\times4\times10^{-6}\times2\times10^{-6}}{0.20\times0.20} N\\\\F=1.8N[/tex]
Since we are calculating the force on the q which is positive, due to Q which is negative. The direction of the force will be towards the right since the electric field for the negative charge will be directed towards the charge, which is towards the right.
So the net force is 1.8N towards right.
Learn more about Coulomb's Law;
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