Respuesta :
To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as
[tex]F=\frac{kq_{1}q_{2} }{r^{2} }[/tex]
Here, [tex]q_{1}[/tex] and[tex]q_{2}[/tex] are the charges on the pith balls, r is the separation between the charges and k is constant and its value is [tex]8.99\times 10^9 N m^2/C^2[/tex].
Given [tex]q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C[/tex] and [tex]r=8 cm=8\times10^{-2} m[/tex].
Substituting these values in above formula we get,
[tex]F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N[/tex]
Thus, the repulsive force between two pith balls is [tex]8.8\times10^{-4}N[/tex].
Solution:
The force between two charges given by Columb's law i[tex]the force between two charges given by Coloumb law is\\ \left | F \right |=\frac{kq_{1}q_{2}}{r^{2}}\\\\ where k is the constant equal to\\ \rn Now,\\ We have \\r=0.008mand q1=q2=-25.0\times10^{-9}\\ \left | F \right |=\left ( 8.099\times10^{9}\frac{Nm^{2}}{c^{2}} \right )\times\left ( \frac{25.0\times10^{-9}\times 25.0\times10^{-9}}{0.008^{2}} \right )\\ =1.21\times10^{-3}[/tex]
Solution:
The force between two charges given by Columb's law i[tex]the force between two charges given by Coloumb law is\\ \left | F \right |=\frac{kq_{1}q_{2}}{r^{2}}\\\\ where k is the constant equal to\\ \rn Now,\\ We have \\r=0.008mand q1=q2=-25.0\times10^{-9}\\ \left | F \right |=\left ( 8.099\times10^{9}\frac{Nm^{2}}{c^{2}} \right )\times\left ( \frac{25.0\times10^{-9}\times 25.0\times10^{-9}}{0.008^{2}} \right )\\ =1.21\times10^{-3}[/tex]
