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

The force between two charges given by Columb's law i
`the force between two charges given by Coloumb law is\\ \left | F \right |=(kq_(1)q_(2))/(r^(2))\\\\ where k is the constant equal to\\  \rn Now,\\ We have \\r=0.008mand q1=q2=-25.0*10^(-9)\\ \left | F \right |=\left ( 8.099*10^(9)(Nm^(2))/(c^(2)) \right )*\left ( (25.0*10^(-9)* 25.0*10^(-9))/(0.008^(2)) \right )\\ =1.21*10^(-3)`

Answer 2

The force between two charges given by Columb's law i
`the force between two charges given by Coloumb law is\\ \left | F \right |=(kq_(1)q_(2))/(r^(2))\\\\ where k is the constant equal to\\  \rn Now,\\ We have \\r=0.008mand q1=q2=-25.0*10^(-9)\\ \left | F \right |=\left ( 8.099*10^(9)(Nm^(2))/(c^(2)) \right )*\left ( (25.0*10^(-9)* 25.0*10^(-9))/(0.008^(2)) \right )\\ =1.21*10^(-3)`

Answer 3

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

`F=(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* 10^9 N m^2/C^2`.

Given
`q_(1) =q_(2) =-25 nC=-25* 10^(-9)C` and
`r=8 cm=8*10^(-2) m`.

Substituting these values in above formula we get,

`F= 8.99* 10^9 N m^2/C^2((-25* 10^(-9)C)(-25* 10^(-9)C))/((8*10^(-2) m)^2) \\\\\ F= 877929.7* 10^(-9)  N\\\\F=8.8*10^(-4)N`

Thus, the repulsive force between two pith balls is
`8.8*10^(-4)N`.