two small pith balls are 1 cm apart in air and carry charges of 2 into

1.	two small pith balls are 1 cm apart in air and carry charges of 2 into 10 power minus 9 coulomb and minus 6 into 10 power minus 9 coulomb respectively calculate the force of attraction if the balls are dust and then separated by a distance of 1 cm what will the force between them
Answer» Given that, Charge on first ball $q=2\times10^{-9}\ C$ Charge on SECOND ball $q=-6\times10^{-9}\ C$ Distance = 1 cm We need to calculate the force of attraction Using formula of force $F=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r^2}$ Put the value into the formula $F=9\times10^{9}\times\dfrac{2\times10^{-9}\times6\times10^{-9}}{(1\times10^{-2})^2}$ $F=0.00108\ N$ $F=1.08\times10^{-3}\ N$ When these TWO charged pith balls are touched then first the opposite charges are NEUTRALISED and then the SEPARATED charge of $-4\times10^{-9}\ C$ is equally DISTRIBUTED on pith balls. We need to calculate the force of repulsion Using formula of force $F=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r^2}$ Put the value into the formula $F=9\times10^{9}\times\dfrac{2\times10^{-9}\times2\times10^{-9}}{(1\times10^{-2})^2}$ $F=0.00036\ N$ $F=3.6\times10^{-4}\ N$ $1.08\times10^{-3}\ N$ Hence, The force of attraction is $3.6\times10^{-4}\ N$ The force of repulsion is

two small pith balls are 1 cm apart in air and carry charges of 2 into 10 power minus 9 coulomb and minus 6 into 10 power minus 9 coulomb respectively calculate the force of attraction if the balls are dust and then separated by a distance of 1 cm what will the force between them

Answer»

Given that,

Charge on first ball $q=2\times10^{-9}\ C$

Charge on SECOND ball $q=-6\times10^{-9}\ C$

Distance = 1 cm

We need to calculate the force of attraction

Using formula of force

$F=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r^2}$

Put the value into the formula

$F=9\times10^{9}\times\dfrac{2\times10^{-9}\times6\times10^{-9}}{(1\times10^{-2})^2}$

$F=0.00108\ N$

$F=1.08\times10^{-3}\ N$

When these TWO charged pith balls are touched then first the opposite charges are NEUTRALISED and then the SEPARATED charge of $-4\times10^{-9}\ C$ is equally DISTRIBUTED on pith balls.