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ABC  is a right angled triangle in which AB = 3 cm  and BC = 4 cm . And \angle ABC=\pi l2. The three charges +15 , +12   and -20 e.s.u.  are placed respectively on A , B  and C . The force acting on B  is

  • Option 1)

    125\; \; dynes\;

  • Option 2)

    \; 35\; \; dynes\;

  • Option 3)

    \; 25\; \; dynes\;

  • Option 4)

    \; Zero

 

Answers (1)

As we learned

Principle of Super Position -

It states that total force acting on a given charge due to number of charges is the Vector sum of the individual force acting on that charge due to all the charges.

- wherein

 

 Net force on     BF_{net}=\sqrt{F_{A}^{2}+F_{C}^{2}}  

\\*F_{A}=\frac{15\times 12}{(3)^{2}}=20\, dyne,\; F_{C}=\frac{12\times 20}{(4)^{2}}=15\, dyne\\*\Rightarrow F_{net}=\sqrt{F_{A}^{2}+F_{C}^{2}}=\sqrt{(20)^{2}+(15)^{2}}=25dyne

 


Option 1)

125\; \; dynes\;

Option 2)

\; 35\; \; dynes\;

Option 3)

\; 25\; \; dynes\;

Option 4)

\; Zero

Posted by

Vakul

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