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Solve it, - Electrostatics - JEE Main-4

Electric charges of 1\mu C,-1\mu C,  and 2\mu C  are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is

  • Option 1)

    0.9 N

  • Option 2)

    1.8 N

  • Option 3)

    2.7 N

  • Option 4)

    3.6 N

 
Answers (1)
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As we learned

Magnitude of the Resultant force -

F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }

- wherein

 

 

F_{A} = force on C due to charge placed at A

=9\times 10^{9}\times \frac{10^{-6}\times2\times10^{-6}}{(10 \times 10^{-2})^{2}}=1.8N

F_{B} = force on C due to charge placed at B

=9\times 10^{9}\times \frac{10^{-6}\times2\times10^{-6}}{(0.1)^{2}}=1.8N      Net force on C

F_{net}=\sqrt{(F_{A})^{2}+(F_{B})^{2}+2F_{A}F_{B}\cos 120^{\circ}}=1.8N

 


Option 1)

0.9 N

Option 2)

1.8 N

Option 3)

2.7 N

Option 4)

3.6 N

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