Question

two small identical spheres having charges 10 microcoulomb and - 90 microcoulomb attract each other with a force of F Newton. if they are kept in contact and then separated by the same distance, the new force between them is

Answer 1

When they are touched charge will be equally distributed.so(10+(-90))/2=-40So new force will be F`=K(-40)^2/r^2Initially the force wasF=10×-90/r^2Deviding both the forcesF`=16F/9

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Answer 2

Answer:

The new force between them is $F'=\dfrac{16}{9}F$.

Explanation:

Given that,

First charge $q_{1}=10\times10^{-6}\ C$

Second charge $q_{2}=-90\times10^{-6}\ C$

Let the distance between the both charges are r.

The force is

$F= \dfrac{kq_{1}q_{2}}{r^2}$

$F= \dfrac{k\times10\times10^{-6}\times(-90\times10^{-6})}{r^2}$

When they are touch then the charge on them will be equal.

The charge will be

$Q= \dfrac{10\times10^{-6}-90\times10^{-6}}{2}$

$Q= 40\times 10^{-6}\ C$

The force is

$F'= \dfrac{kq^2}{r^2}$

$F'= \dfrac{k\times(-40\toimes10^{-6})^2}{r^2}$

The ratio of F' and F is

$\dfrac{F'}{F}=\dfrac{\dfrac{k\times(40\times10^{-6})^2}{r^2}}{\dfrac{k\times10\times10^{-6}\times(90\times10^{-6})}{r^2}}$

$\dfrac{F'}{F}=\dfrac{16}{9}$

$F'=\dfrac{16}{9}F$

Hence, The new force between them is $F'=\dfrac{16}{9}F$.