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Two point charges +3\mu C and +8\mu C repel each other with a force of 40N. If a charge of -5\mu Cis added to each of them, then the force between them will become

  • Option 1)

    -10N\; \;

  • Option 2)

    \; \; +10N\; \;

  • Option 3)

    \; +20N\;

  • Option 4)

    \; -20N

 

Answers (1)

As we learned

By conduction -

When two conducters brought in contact.

- wherein

i.e The charge will spread over both the conducters.

 

 In second case, charges will be  -2\mu C\; and\; +3\mu C\;

Since\; \; F\alpha Q_{1}Q_{2}\, i.e\: \frac{F}{F'}=\frac{Q_{1}Q_{2}}{Q'_{1}Q'_{2}}

\therefore \frac{40}{F'}=\frac{3\times 8}{-2\times 3}=-4\Rightarrow F'=10N\; \; (Attractive)

 


Option 1)

-10N\; \;

Option 2)

\; \; +10N\; \;

Option 3)

\; +20N\;

Option 4)

\; -20N

Posted by

Vakul

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