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Need explanation for: - Electrostatics - JEE Main-5

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)

    0.9N\; \;

  • Option 2)

    \; 1.8N\; \;

  • Option 3)

    \; 2.7N\;

  • Option 4)

    \; 3.6N

 
Answers (1)
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As we learned

By conduction -

When two conducters brought in contact.

- wherein

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

 

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

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

FB = 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)

0.9N\; \;

Option 2)

\; 1.8N\; \;

Option 3)

\; 2.7N\;

Option 4)

\; 3.6N

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