Question

Two point charges 5 micro coulomb and 10 micro coulomb are separated by a distance 'r' in air. If an additional charge of -4 micro coulomb is given to each, by what factor does the force between the charges change?

Answer 1

Answer: 3/25

Explanation:

Answer 2

The new electric force between charges becomes 0.12 time the initial force.

Explanation:

It is given that,

Charge 1, $q_1=5\ \mu C=5\times 10^{-6}\ C$

Charge 2, $q_2=10\ \mu C=10^{-5}\ C$

The electric force between charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=k\dfrac{5\times 10^{-6}\times 10^{-5}}{r^2}$........(1)

If an additional charge of -4 micro coulomb is given to each, new charges becomes,

Charge 1, $q_1=1\ \mu C=10^{-6}\ C$

Charge 2, $q_2=6\ \mu C=6\times 10^{-6}\ C$

New electric force is given by,

$F'=k\dfrac{10^{-6}\times 6\times 10^{-6}}{r^2}$..........(2)

Dividing equation (1) and (2) we get :

$\dfrac{F}{F'}=\dfrac{k\dfrac{5\times 10^{-6}\times 10^{-5}}{r^2}}{k\dfrac{10^{-6}\times 6\times 10^{-6}}{r^2}}$

$\dfrac{F}{F'}=8.34$

$F'=0.12F$

So, the new electric force between charges becomes 0.12 time the initial force. Hence, this is the required solution.

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