Question

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Answer 1

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2}  }{r^{2} }$

Here, $q_{1}$ and$q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  $8.99\times 10^9 N m^2/C^2$.

Given  $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and  $r=8 cm=8\times10^{-2} m$.

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N$

Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$.

Answer 2

Solution:

The force between two charges given by Columb's law i$the force between two charges given by Coloumb law is\\ \left | F \right |=\frac{kq_{1}q_{2}}{r^{2}}\\\\ where k is the constant equal to\\  \rn Now,\\ We have \\r=0.008mand q1=q2=-25.0\times10^{-9}\\ \left | F \right |=\left ( 8.099\times10^{9}\frac{Nm^{2}}{c^{2}} \right )\times\left ( \frac{25.0\times10^{-9}\times 25.0\times10^{-9}}{0.008^{2}} \right )\\ =1.21\times10^{-3}$

Answer 3

Solution:

The force between two charges given by Columb's law i$the force between two charges given by Coloumb law is\\ \left | F \right |=\frac{kq_{1}q_{2}}{r^{2}}\\\\ where k is the constant equal to\\  \rn Now,\\ We have \\r=0.008mand q1=q2=-25.0\times10^{-9}\\ \left | F \right |=\left ( 8.099\times10^{9}\frac{Nm^{2}}{c^{2}} \right )\times\left ( \frac{25.0\times10^{-9}\times 25.0\times10^{-9}}{0.008^{2}} \right )\\ =1.21\times10^{-3}$