Question

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 1

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

$F=9\times10^{9}\times\dfrac{2\times10^{-9}\times2\times10^{-9}}{(1\times10^{-2})^2}$

$F=0.00036\ N$

$F=3.6\times10^{-4}\ N$

Hence, The force of attraction is $1.08\times10^{-3}\ N$

The force of repulsion is $3.6\times10^{-4}\ N$

Answer 2

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

$F=9\times10^{9}\times\dfrac{2\times10^{-9}\times2\times10^{-9}}{(1\times10^{-2})^2}$

$F=0.00036\ N$

$F=3.6\times10^{-4}\ N$

Hence, The force of attraction is $1.08\times10^{-3}\ N$

The force of repulsion is $3.6\times10^{-4}\ N$

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