Answer: [tex]247.12\ N[/tex]
Explanation:
Given
Magnitude of the charges
[tex]q_1=4.47\times 10^{-6}\ C[/tex]
[tex]q_2=1.86\times 10^{-6}\ C[/tex]
Distance between them [tex]d=17.4\ mm[/tex]
As both charges are of same sign, they must repel each other
Force experienced by second charge is
[tex]\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N[/tex]
Thus, charge 2 experience a force of [tex]247.12\ N[/tex]
Answer:
The force between the two charges is 247.15 N.
Explanation:
Charge, q = 4.47 x 10^-6 C
charge, q' = 1.86 x 10^-6 C
distance, d = 17.4 mm
Let the force is F.
The force is given by the Coulomb's law:
[tex]F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N[/tex]
